Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is I. If the same rod is bent into a ring and its moment of inertia about its diameter is I'. Then the ratio I/I' will be
🥳 Wohoo! Correct answer
I = mr²/k
Rod: I = mL²/12 Ring: I' = mR²/2 = m(L/2π)²/2
Substitute R = L/2π
Simplify to get I/I' = 2π²/3
Wrong substitution of radius
😢 Uh oh! Incorrect answer, Try again
Compare MI for different shapes
I = mr²/k
Rod: I = mL²/12 Ring: I' = mR²/2 = m(L/2π)²/2
Substitute R = L/2π
Simplify to get I/I' = 2π²/3
Wrong substitution of radius
If the mass of a body is M on the surface of the earth, the mass of the same body on the surface of the moon will be
🥳 Wohoo! Correct answer
m = constant, W = mg
Mass is an intrinsic property
Mass doesn't change with location
Mass remains M regardless of gravity
Confusing mass and weight
😢 Uh oh! Incorrect answer, Try again
Mass vs weight distinction
m = constant, W = mg
Mass is an intrinsic property
Mass doesn't change with location
Mass remains M regardless of gravity
Confusing mass and weight
The average power dissipated in a pure inductor is
🥳 Wohoo! Correct answer
P = VI cos φ
For pure inductor current lags voltage by 90°
Average power = VI cos φ
When φ = 90°, cos φ = 0, so P = 0
Forgetting phase angle
😢 Uh oh! Incorrect answer, Try again
Consider phase difference
P = VI cos φ
For pure inductor current lags voltage by 90°
Average power = VI cos φ
When φ = 90°, cos φ = 0, so P = 0
Forgetting phase angle
A transformer is used to light 100W-110V lamp from 220V mains. If the main current is 0.5A, the efficiency of the transformer is
🥳 Wohoo! Correct answer
η = (P_out/P_in)×100
Output power = 100W Input power = 220V×0.5A = 110W
Efficiency = (Output/Input)×100
is equal to (100/110)×100 = 90.9% ≈ 90%
Not considering power ratio
😢 Uh oh! Incorrect answer, Try again
Compare input/output powers
η = (P_out/P_in)×100
Output power = 100W Input power = 220V×0.5A = 110W
Efficiency = (Output/Input)×100
is equal to (100/110)×100 = 90.9% ≈ 90%
Not considering power ratio
In an LCR circuit, at resonance
🥳 Wohoo! Correct answer
Z = √(R² + (XL-XC)²)
At resonance XL = XC
Net reactance is zero
Therefore current and voltage are in phase
Confusing phase relations
😢 Uh oh! Incorrect answer, Try again
Consider reactance cancellation
Z = √(R² + (XL-XC)²)
At resonance XL = XC
Net reactance is zero
Therefore current and voltage are in phase
Confusing phase relations
A galvanometer of resistance 50Ω gives a full-scale deflection for a current 5×10⁻⁴ A. The resistance that should be connected in series with the galvanometer to read 3V is
🥳 Wohoo! Correct answer
R = V/I - Rg
Use Ohm's law: R = V/I where V = 3V, I = 5×10⁻⁴ A
Total R = 3/(5×10⁻⁴) = 6000Ω
Required R = 6000 - 50 = 5950Ω
Forgetting galvanometer resistance
😢 Uh oh! Incorrect answer, Try again
Consider existing resistance
R = V/I - Rg
Use Ohm's law: R = V/I where V = 3V, I = 5×10⁻⁴ A
Total R = 3/(5×10⁻⁴) = 6000Ω
Required R = 6000 - 50 = 5950Ω
Forgetting galvanometer resistance
The resistance of the bulb filament is 100Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200Ω at a temperature
🥳 Wohoo! Correct answer
R = R₀(1 + αΔT)
Use R = R₀[1 + α(T-T₀)] 200 = 100[1 + 0.005(T-100)]
2 = 1 + 0.005(T-100) 1 = 0.005(T-100)
T = 100 + 200 = 400°C
Wrong temperature reference
😢 Uh oh! Incorrect answer, Try again
Consider reference temperature
R = R₀(1 + αΔT)
Use R = R₀[1 + α(T-T₀)] 200 = 100[1 + 0.005(T-100)]
2 = 1 + 0.005(T-100) 1 = 0.005(T-100)
T = 100 + 200 = 400°C
Wrong temperature reference
In Wheatstone's network P=2Ω, Q=2Ω, R=2Ω and S=3Ω. The resistance with which S is to be shunted in order that the bridge may be balanced is
🥳 Wohoo! Correct answer
R₁R₄ = R₂R₃
For balance: P/Q = R/S₁ where S₁ is effective resistance
With x in parallel with S: S₁ = (S×x)/(S+x) = (3x)/(3+x)
Solving 2/2 = 2/S₁ gives x = 6Ω
Wrong parallel combination
😢 Uh oh! Incorrect answer, Try again
Use balanced bridge condition
R₁R₄ = R₂R₃
For balance: P/Q = R/S₁ where S₁ is effective resistance
With x in parallel with S: S₁ = (S×x)/(S+x) = (3x)/(3+x)
Solving 2/2 = 2/S₁ gives x = 6Ω
Wrong parallel combination
Three resistances 2Ω, 3Ω and 4Ω are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be
🥳 Wohoo! Correct answer
I = V/R, V = same
In parallel, voltage V is same across all resistors Using I = V/R For 2Ω: I₁ = V/2
For 3Ω: I₂ = V/3 For 4Ω: I₃ = V/4
Ratio I₁:I₂:I₃ = V/2:V/3:V/4 = 6:4:3
Not using reciprocal resistance
😢 Uh oh! Incorrect answer, Try again
Same voltage in parallel
I = V/R, V = same
In parallel, voltage V is same across all resistors Using I = V/R For 2Ω: I₁ = V/2
For 3Ω: I₂ = V/3 For 4Ω: I₃ = V/4
Ratio I₁:I₂:I₃ = V/2:V/3:V/4 = 6:4:3
Not using reciprocal resistance
Four identical cells of emf E and internal resistance r are to be connected in series. Suppose if one of the cells is connected wrongly. The equivalent emf and effective internal resistance of the combination is
🥳 Wohoo! Correct answer
Eₑq = ΣE, rₑq = Σr
For three cells in same direction: 3E For one cell in opposite: -E Net EMF = 3E - E = 2E
Internal resistance adds regardless of cell direction r₁ + r₂ + r₃ + r₄ = 4r
Therefore equivalent EMF = 2E and total internal resistance = 4r
Not considering cell direction
😢 Uh oh! Incorrect answer, Try again
Direction affects EMF but not resistance
Eₑq = ΣE, rₑq = Σr
For three cells in same direction: 3E For one cell in opposite: -E Net EMF = 3E - E = 2E
Internal resistance adds regardless of cell direction r₁ + r₂ + r₃ + r₄ = 4r
Therefore equivalent EMF = 2E and total internal resistance = 4r
Not considering cell direction
In a series L.C.R. circuit an alternating emf (v) and current (i) are given by the equation v = v₀ sin ωt, i = i₀ sin (ωt + π/3). The average power dissipated in the circuit over a cycle of AC is
🥳 Wohoo! Correct answer
P = VIcosφ
Power P = VI cosφ where φ is phase difference
Phase difference = π/3, so cosφ = 1/2
Average power = (V₀I₀/2)×(1/2) = V₀I₀/4
Students might ignore phase factor
😢 Uh oh! Incorrect answer, Try again
Consider phase difference effect
P = VIcosφ
Power P = VI cosφ where φ is phase difference
Phase difference = π/3, so cosφ = 1/2
Average power = (V₀I₀/2)×(1/2) = V₀I₀/4
Students might ignore phase factor
In a series L.C.R circuit, the potential drop across L, C and R respectively are 40 V, 120 V and 60 V. Then the source voltage is
🥳 Wohoo! Correct answer
Vs = √[(VC-VL)² + VR²]
Apply vector addition since voltages are phase shifted
Calculate source voltage: Vs = √[(VC-VL)² + VR²]
Substitute: Vs = √[(120-40)² + 60²] = √(6400+3600) = 100V
Students might add voltages directly
😢 Uh oh! Incorrect answer, Try again
Inductive and capacitive voltages subtract
Vs = √[(VC-VL)² + VR²]
Apply vector addition since voltages are phase shifted
Calculate source voltage: Vs = √[(VC-VL)² + VR²]
Substitute: Vs = √[(120-40)² + 60²] = √(6400+3600) = 100V
Students might add voltages directly
A capacitor of capacitance 10 μF is connected to an AC source and an AC ammeter. If the source voltage varies as V = 50√2 sin 100t, the reading of the ammeter is
🥳 Wohoo! Correct answer
XC = 1/ωC, I = V/XC
Calculate impedance of capacitor: XC = 1/ωC
Find RMS voltage: Vrms = 50V
Calculate current: Irms = Vrms/XC = 50mA
Students might use peak values
😢 Uh oh! Incorrect answer, Try again
Use capacitive reactance formula
XC = 1/ωC, I = V/XC
Calculate impedance of capacitor: XC = 1/ωC
Find RMS voltage: Vrms = 50V
Calculate current: Irms = Vrms/XC = 50mA
Students might use peak values
The rms value of current in a 50 Hz AC circuit is 6 A. The average value of AC current over a cycle is
🥳 Wohoo! Correct answer
Irms = I₀/√2
Consider complete AC cycle
Note current is sinusoidal and symmetric
Average of complete symmetric cycle is zero
Students might confuse with peak value
😢 Uh oh! Incorrect answer, Try again
Consider symmetry of AC wave
Irms = I₀/√2
Consider complete AC cycle
Note current is sinusoidal and symmetric
Average of complete symmetric cycle is zero
Students might confuse with peak value
A galvanometer coil has a resistance of 50 Ω and the meter shows full-scale deflection for a current of 5 mA. This galvanometer is converted into a voltmeter of range 0-20 V by connecting
🥳 Wohoo! Correct answer
R = V/I
Use Ohm's law: V = IR where I = 5mA = 0.005A
Calculate required total resistance: R = V/I = 20/0.005 = 4000Ω
Required series resistance = 4000 - 50 = 3950Ω
Students might use parallel combination
😢 Uh oh! Incorrect answer, Try again
Series resistance needed for voltage range
R = V/I
Use Ohm's law: V = IR where I = 5mA = 0.005A
Calculate required total resistance: R = V/I = 20/0.005 = 4000Ω
Required series resistance = 4000 - 50 = 3950Ω
Students might use parallel combination
Two heating coils of resistances 10 Ω and 20 Ω are connected in parallel and connected to a battery of emf 12V and internal resistance 1Ω. The power consumed by them are in the ratio
🥳 Wohoo! Correct answer
P = V²/R
Calculate potential difference across parallel combination
Find currents through each resistor: I₁ = V/R₁, I₂ = V/R₂
Power ratio = V²/R₁ : V²/R₂ = R₂/R₁ = 2:1
Students might use current ratio
😢 Uh oh! Incorrect answer, Try again
Compare power using resistance ratio
P = V²/R
Calculate potential difference across parallel combination
Find currents through each resistor: I₁ = V/R₁, I₂ = V/R₂
Power ratio = V²/R₁ : V²/R₂ = R₂/R₁ = 2:1
Students might use current ratio
The quantity of a charge that will be transferred by a current flow of 20 A over 1 hour 30 minutes period is
🥳 Wohoo! Correct answer
Q = It
Convert time to seconds: 1h 30min = 5400s
Apply Q = It: Q = 20 × 5400
Calculate: Q = 108000 = 10.8×10⁴ C
Students might use wrong time units
😢 Uh oh! Incorrect answer, Try again
Remember to convert time to seconds
Q = It
Convert time to seconds: 1h 30min = 5400s
Apply Q = It: Q = 20 × 5400
Calculate: Q = 108000 = 10.8×10⁴ C
Students might use wrong time units
Mobility of free electrons in a conductor is
🥳 Wohoo! Correct answer
μ = qτ/m
Define mobility μ = qτ/m where τ is relaxation time
Observe direct relationship between μ and τ
Mobility increases linearly with relaxation time
Students might confuse with conductivity
😢 Uh oh! Incorrect answer, Try again
Consider charge carrier motion
μ = qτ/m
Define mobility μ = qτ/m where τ is relaxation time
Observe direct relationship between μ and τ
Mobility increases linearly with relaxation time
Students might confuse with conductivity
A coil of inductive reactance (1/√3)Ω and resistance 1Ω is connected to a 200 V, 50 Hz A.C. supply. The time lag between the maximum voltage and current is
🥳 Wohoo! Correct answer
φ = tan⁻¹(X_L/R)
Calculate phase angle: tanφ = X_L/R = 1/√3
Find φ = 30°
Time lag = φ/2πf = 30°/(2π×50) = 1/600 s
Not converting angle to time
😢 Uh oh! Incorrect answer, Try again
Phase difference determines time lag
φ = tan⁻¹(X_L/R)
Calculate phase angle: tanφ = X_L/R = 1/√3
Find φ = 30°
Time lag = φ/2πf = 30°/(2π×50) = 1/600 s
Not converting angle to time
The output of a step-down transformer is measured to be 48 V when connected to a 12 w bulb. The value of peak current is
🥳 Wohoo! Correct answer
I_peak = I_rms × √2
Calculate RMS current: I_rms = P/V = 12/48 = 0.25 A
Peak current = I_rms × √2
I_peak = 0.25 × √2 = 1/(2√2) A
Confusing RMS and peak values
😢 Uh oh! Incorrect answer, Try again
RMS values related to peak by √2
I_peak = I_rms × √2
Calculate RMS current: I_rms = P/V = 12/48 = 0.25 A
Peak current = I_rms × √2
I_peak = 0.25 × √2 = 1/(2√2) A
Confusing RMS and peak values
A galvanometer of resistance 50Ω is connected to a battery of 3V along with a resistance of 2950Ω in series shows full-scale deflection of 30 divisions. The additional series resistance required to reduce the deflection to 20 divisions is
🥳 Wohoo! Correct answer
I = V/R
Current for 30 divisions: I = 3/(3000) A
For 20 divisions: I' = (20/30)×I
Calculate required total resistance: R = 3/I' - 50 = 4440Ω
Not considering proportionality
😢 Uh oh! Incorrect answer, Try again
Deflection proportional to current
I = V/R
Current for 30 divisions: I = 3/(3000) A
For 20 divisions: I' = (20/30)×I
Calculate required total resistance: R = 3/I' - 50 = 4440Ω
Not considering proportionality
In metre bridge experiment, with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap, the balancing length obtained is 'l'. If the temperature of the water is increased, the new balancing length is
🥳 Wohoo! Correct answer
R = R₀(1 + αΔT)
Resistance increases with temperature for metals
Balance condition: R₁/R₂ = l₁/l₂
As R₁ increases, l₁ must increase
Not considering temperature effect on resistance
😢 Uh oh! Incorrect answer, Try again
Remember temperature coefficient is positive for metals
R = R₀(1 + αΔT)
Resistance increases with temperature for metals
Balance condition: R₁/R₂ = l₁/l₂
As R₁ increases, l₁ must increase
Not considering temperature effect on resistance
A piece of copper is to be shaped into a conducting wire of maximum resistance. The suitable length and diameter are
🥳 Wohoo! Correct answer
R = ρL/A
Resistance formula: R = ρL/A, where A = πd²/4
Resistance ∝ L and ∝ 1/d²
Maximum R needs maximum L and minimum d
Not considering area dependence
😢 Uh oh! Incorrect answer, Try again
Resistance inversely proportional to cross-sectional area
R = ρL/A
Resistance formula: R = ρL/A, where A = πd²/4
Resistance ∝ L and ∝ 1/d²
Maximum R needs maximum L and minimum d
Not considering area dependence
A cylindrical conductor of diameter 0.1 mm carries a current of 90 mA. The current density (in Am⁻²) is (π ≃ 3)
🥳 Wohoo! Correct answer
J = I/A
Calculate area: A = πr² = π(0.05×10⁻³)² m²
Convert current to amperes: I = 90×10⁻³ A
Calculate J = I/A = 1.2×10⁷ A/m²
Forgetting to convert units to SI
😢 Uh oh! Incorrect answer, Try again
Remember to convert all units to SI
J = I/A
Calculate area: A = πr² = π(0.05×10⁻³)² m²
Convert current to amperes: I = 90×10⁻³ A
Calculate J = I/A = 1.2×10⁷ A/m²
Forgetting to convert units to SI
A step-up transformer operates on a 230 V line and loads current of 2 A. The ratio of primary and secondary windings is 1:25. Then the current in the primary is
🥳 Wohoo! Correct answer
N_p/N_s = I_s/I_p = V_s/V_p
Use transformer ratio equation: N_p/N_s = I_s/I_p
Given N_p:N_s = 1:25 and I_s = 2A
Calculate I_p = 2 × 25 = 50A
Not applying ratio correctly
😢 Uh oh! Incorrect answer, Try again
Consider power conservation
N_p/N_s = I_s/I_p = V_s/V_p
Use transformer ratio equation: N_p/N_s = I_s/I_p
Given N_p:N_s = 1:25 and I_s = 2A
Calculate I_p = 2 × 25 = 50A
Not applying ratio correctly
In Karnataka, the normal domestic power supply AC is 220 V, 50 Hz. Here 220 V and 50 Hz refer to
🥳 Wohoo! Correct answer
V_RMS = V_peak/√2
Understand difference between peak and RMS values
Note that quoted AC values are typically RMS
Remember V_RMS = V_peak/√2
Confusing peak and RMS
😢 Uh oh! Incorrect answer, Try again
Consider standard power specifications
V_RMS = V_peak/√2
Understand difference between peak and RMS values
Note that quoted AC values are typically RMS
Remember V_RMS = V_peak/√2
Confusing peak and RMS
In the series LCR circuit, the power dissipation is through
🥳 Wohoo! Correct answer
P = I²R
Consider energy storage in L and C
Note that only R dissipates energy
L and C store energy but don't dissipate
Confusing storage with dissipation
😢 Uh oh! Incorrect answer, Try again
Think about phase relationships
P = I²R
Consider energy storage in L and C
Note that only R dissipates energy
L and C store energy but don't dissipate
Confusing storage with dissipation
Two cells of internal resistances r₁ and r₂ and of the same emf are connected in series, across a resistor of resistance R. If the terminal potential difference across the cells of internal resistance r₁ is zero, then the value of R is
🥳 Wohoo! Correct answer
V = E - Ir
Write Kirchhoff's voltage law equation for the circuit
Apply condition that voltage across r₁ is zero
Solve for R using these conditions
Not applying KVL correctly
😢 Uh oh! Incorrect answer, Try again
Consider voltage drops across each element
V = E - Ir
Write Kirchhoff's voltage law equation for the circuit
Apply condition that voltage across r₁ is zero
Solve for R using these conditions
Not applying KVL correctly
A 100 W bulb is connected to an AC source of 220 V, 50 Hz. Then the current flowing through the bulb is
🥳 Wohoo! Correct answer
P = VI
Use power formula: P = VI for AC circuit
Calculate current: I = P/V = 100/220
Simplify to get 5/11 A
Not using RMS values
😢 Uh oh! Incorrect answer, Try again
Consider RMS values for AC
P = VI
Use power formula: P = VI for AC circuit
Calculate current: I = P/V = 100/220
Simplify to get 5/11 A
Not using RMS values
If the last band on the carbon resistor is absent, then the tolerance is
🥳 Wohoo! Correct answer
None
Recall color code rules for resistors
Note that absence of tolerance band has specific meaning
Absence of tolerance band indicates 20% tolerance
Assuming zero tolerance
😢 Uh oh! Incorrect answer, Try again
Remember default tolerance value
None
Recall color code rules for resistors
Note that absence of tolerance band has specific meaning
Absence of tolerance band indicates 20% tolerance
Assuming zero tolerance
Ohm's Law is applicable to
🥳 Wohoo! Correct answer
V = IR
Understand that Ohm's law states V = IR with R constant
Check which device maintains constant resistance independent of voltage
Among options, only conductors maintain linear V-I relationship
Applying to semiconductors
😢 Uh oh! Incorrect answer, Try again
Look for linear V-I characteristic
V = IR
Understand that Ohm's law states V = IR with R constant
Check which device maintains constant resistance independent of voltage
Among options, only conductors maintain linear V-I relationship
Applying to semiconductors
In a potentiometer experiment, the balancing point with a cell is at a length 240cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120cm. The internal resistance of the cell is
🥳 Wohoo! Correct answer
l₁/l₂ = (E/(R+r))/(E/(R+r+r'))
Balance length ∝ terminal potential
V₁/V₂ = l₁/l₂ = 2
Calculate r = 2Ω from ratio
Not using correct resistance ratio
😢 Uh oh! Incorrect answer, Try again
Use potential ratio
l₁/l₂ = (E/(R+r))/(E/(R+r+r'))
Balance length ∝ terminal potential
V₁/V₂ = l₁/l₂ = 2
Calculate r = 2Ω from ratio
Not using correct resistance ratio
Kirchhoff's junction rule is a reflection of
🥳 Wohoo! Correct answer
ΣI = 0 at junction
At junction, total incoming current equals outgoing
This represents charge conservation
No charge accumulation at junction
Not relating to charge conservation
😢 Uh oh! Incorrect answer, Try again
Think about charge flow
ΣI = 0 at junction
At junction, total incoming current equals outgoing
This represents charge conservation
No charge accumulation at junction
Not relating to charge conservation
An inductor of inductance L and resistor R are joined together in series and connected by a source of frequency ω. The power dissipated in the circuit is
🥳 Wohoo! Correct answer
P = I²R = V²R/Z²
Total impedance Z = √(R² + ω²L²)
Current I = V/Z
Power P = I²R = (V²/Z²)R = V²R/(R² + ω²L²)
Not using impedance correctly
😢 Uh oh! Incorrect answer, Try again
Use power formula with impedance
P = I²R = V²R/Z²
Total impedance Z = √(R² + ω²L²)
Current I = V/Z
Power P = I²R = (V²/Z²)R = V²R/(R² + ω²L²)
Not using impedance correctly
Masses of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance are
🥳 Wohoo! Correct answer
R = ρl/A, A = m/ρl
R = ρl/A, where A = m/ρl
R ∝ l²/m (substituting A)
Calculate ratios: (5²/1):(3²/3):(1²/5) = 125:15:1
Not relating mass to area
😢 Uh oh! Incorrect answer, Try again
Use resistance and density formulas
R = ρl/A, A = m/ρl
R = ρl/A, where A = m/ρl
R ∝ l²/m (substituting A)
Calculate ratios: (5²/1):(3²/3):(1²/5) = 125:15:1
Not relating mass to area
The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value?
🥳 Wohoo! Correct answer
T = 1/f, ω = 2πf
For AC, i = I₀sin(ωt), Irms = I₀/√2
Time to reach peak from rms means sin(ωt) changes from 1/√2 to 1
t = [sin⁻¹(1) - sin⁻¹(1/√2)]/ω = 2.5ms
Not considering shortest time interval
😢 Uh oh! Incorrect answer, Try again
Consider time for 1/√2 to 1 in sine function
T = 1/f, ω = 2πf
For AC, i = I₀sin(ωt), Irms = I₀/√2
Time to reach peak from rms means sin(ωt) changes from 1/√2 to 1
t = [sin⁻¹(1) - sin⁻¹(1/√2)]/ω = 2.5ms
Not considering shortest time interval
Though the electron drift velocity is small and electron charge is very small, a conductor can carry an appreciably large current because
🥳 Wohoo! Correct answer
I = neAvd
Current I = neAvd where n is number density
n ≈ 10²⁸ electrons/m³ compensates for small vd
Large n makes current significant
Not understanding current mechanism
😢 Uh oh! Incorrect answer, Try again
Consider current formula
I = neAvd
Current I = neAvd where n is number density
n ≈ 10²⁸ electrons/m³ compensates for small vd
Large n makes current significant
Not understanding current mechanism
The colour code for a carbon resistor of resistance 0.28k+10% is
🥳 Wohoo! Correct answer
Standard resistor color code
First convert 0.28k to 280Ω. For 4-band code: 1st band gives first digit, 2nd band gives second digit, 3rd band is multiplier, 4th band is tolerance.
Red = 2 (1st digit), Grey = 8 (2nd digit), Brown = ×10¹ (multiplier), Silver = ±10% (tolerance).
These bands in sequence give 280Ω ±10%, which equals 0.28kΩ ±10%
Confusion in color code sequence
😢 Uh oh! Incorrect answer, Try again
Remember color code sequence: digit, digit, multiplier, tolerance
Standard resistor color code
First convert 0.28k to 280Ω. For 4-band code: 1st band gives first digit, 2nd band gives second digit, 3rd band is multiplier, 4th band is tolerance.
Red = 2 (1st digit), Grey = 8 (2nd digit), Brown = ×10¹ (multiplier), Silver = ±10% (tolerance).
These bands in sequence give 280Ω ±10%, which equals 0.28kΩ ±10%
Confusion in color code sequence
The power factor of R-L circuit is 1/3. If the inductive reactance is 2Ω. The value of resistance is
🥳 Wohoo! Correct answer
cos θ = R/Z, tan θ = Xₗ/R
Power factor = cos θ = R/Z = 1/3, where Z is impedance. Also know that tan θ = Xₗ/R, where Xₗ is inductive reactance = 2Ω.
From power factor = 1/3, find θ. Then use tan θ = 2/R to find R.
Since cos θ = 1/3, tan θ = 2/R = √8. Therefore R = √2 Ω
Confusion between power factor and phase angle
😢 Uh oh! Incorrect answer, Try again
Use power factor and reactance relationships
cos θ = R/Z, tan θ = Xₗ/R
Power factor = cos θ = R/Z = 1/3, where Z is impedance. Also know that tan θ = Xₗ/R, where Xₗ is inductive reactance = 2Ω.
From power factor = 1/3, find θ. Then use tan θ = 2/R to find R.
Since cos θ = 1/3, tan θ = 2/R = √8. Therefore R = √2 Ω
Confusion between power factor and phase angle
A potentiometer has a uniform wire of length 5m. A battery of emf 10V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200 cm. The emf of the secondary cell is
🥳 Wohoo! Correct answer
E₂/E₁ = l₂/l₁
In a potentiometer, voltage varies linearly with length. First calculate potential gradient: 10V/5m = 2V/m. This tells us voltage drop per meter of wire.
At balance point, potential drop across wire equals emf of secondary cell. Balance length is 200cm = 2m.
Calculate emf of secondary cell: (2V/m)(2m) = 4V. This is the potential difference needed to balance secondary cell emf.
Not using proper length ratio
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Potential varies linearly along potentiometer wire
E₂/E₁ = l₂/l₁
In a potentiometer, voltage varies linearly with length. First calculate potential gradient: 10V/5m = 2V/m. This tells us voltage drop per meter of wire.
At balance point, potential drop across wire equals emf of secondary cell. Balance length is 200cm = 2m.
Calculate emf of secondary cell: (2V/m)(2m) = 4V. This is the potential difference needed to balance secondary cell emf.
Not using proper length ratio
A car has a fresh storage battery of e.m.f 12V and internal resistance 2×10⁻² Ω. If the starter motor draws a current of 80A. Then the terminal voltage when the starter is on is
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V = E - Ir
Start with the basic equation for terminal voltage: V = E - Ir, where E is emf, I is current drawn, and r is internal resistance. The voltage drops due to internal resistance when current flows.
Carefully substitute the given values: V = 12V - (80A)(2×10⁻² Ω). The second term represents voltage drop across internal resistance.
Calculate: V = 12 - 1.6 = 10.4V. This shows how terminal voltage is less than emf due to internal voltage drop.
Not understanding voltage drop concept
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Terminal voltage is always less than emf when current flows
V = E - Ir
Start with the basic equation for terminal voltage: V = E - Ir, where E is emf, I is current drawn, and r is internal resistance. The voltage drops due to internal resistance when current flows.
Carefully substitute the given values: V = 12V - (80A)(2×10⁻² Ω). The second term represents voltage drop across internal resistance.
Calculate: V = 12 - 1.6 = 10.4V. This shows how terminal voltage is less than emf due to internal voltage drop.
Not understanding voltage drop concept
For conductor with potential V, charge q, time t, current I: Which statements true? (i)KE increases by IVt (ii)PE decreases by IVt (iii)Thermal energy increases by IVt
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ΔKE = -ΔPE = qV
When charge moves through potential difference: It gains KE = qV = IVt
It loses PE by same amount = IVt
Without collisions, KE gain equals PE loss. With collisions, energy becomes thermal
Energy conversion confusion
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Energy conversion in circuit
ΔKE = -ΔPE = qV
When charge moves through potential difference: It gains KE = qV = IVt
It loses PE by same amount = IVt
Without collisions, KE gain equals PE loss. With collisions, energy becomes thermal
Energy conversion confusion
A metal rod of length 10 cm and a rectangular cross-section of 1cm × 1/2cm is connected to battery across opposite faces. The resistance will be
🥳 Wohoo! Correct answer
R = ρl/A
Understand that resistance R = ρl/A, where l is length of current path and A is cross-sectional area perpendicular to current. When battery is connected to 1cm × 1/2cm face, current travels 10cm length through smallest area.
Compare all possible configurations: For 1cm × 1/2cm faces: l = 10cm, A = 0.5cm². For 10cm × 1/2cm faces: l = 1cm, A = 5cm². For 10cm × 1cm faces: l = 1/2cm, A = 10cm².
Calculate R ∝ l/A for each case. The first case gives largest l/A ratio, hence maximum resistance.
Not considering all possible current paths
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Resistance is maximum when current path is longest and area is smallest
R = ρl/A
Understand that resistance R = ρl/A, where l is length of current path and A is cross-sectional area perpendicular to current. When battery is connected to 1cm × 1/2cm face, current travels 10cm length through smallest area.
Compare all possible configurations: For 1cm × 1/2cm faces: l = 10cm, A = 0.5cm². For 10cm × 1/2cm faces: l = 1cm, A = 5cm². For 10cm × 1cm faces: l = 1/2cm, A = 10cm².
Calculate R ∝ l/A for each case. The first case gives largest l/A ratio, hence maximum resistance.
Not considering all possible current paths
A copper wire of length 1m, area 5×10⁻⁷m², current 1A has 8×10²⁸ free electrons per m³. Electron drift time from one end to other is
🥳 Wohoo! Correct answer
vd = I/neA
First find drift velocity: I = neAvd, where n is electron density
vd = I/(neA) = 1/(8×10²⁸×1.6×10⁻¹⁹×5×10⁻⁷)
Time = Length/velocity = 1/vd = 6.4×10³s
Unit conversion errors
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Drift velocity is very small
vd = I/neA
First find drift velocity: I = neAvd, where n is electron density
vd = I/(neA) = 1/(8×10²⁸×1.6×10⁻¹⁹×5×10⁻⁷)
Time = Length/velocity = 1/vd = 6.4×10³s
Unit conversion errors
Wire of resistance 3Ω stretched to twice length. New resistance is
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R ∝ l²/V
Use R = ρl/A. When stretched, l → 2l
By volume conservation, A₁l₁ = A₂l₂, so A → A/2
Therefore R → 4R = 12Ω
Area change forgotten
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Volume remains constant
R ∝ l²/V
Use R = ρl/A. When stretched, l → 2l
By volume conservation, A₁l₁ = A₂l₂, so A → A/2
Therefore R → 4R = 12Ω
Area change forgotten
LCR circuit: R=300Ω, L=0.9H, C=2.0μF, ω=1000rad/s. Impedance is:
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Z = √(R² + (XL - XC)²)
Impedance Z = √(R² + (XL - XC)²). Calculate XL = ωL and XC = 1/ωC separately.
XL = 1000×0.9 = 900Ω, XC = 1/(1000×2×10⁻⁶) = 500Ω
Z = √(300² + (900-500)²) = √(90000 + 160000) = 500Ω
Adding instead of subtracting reactances
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Remember to subtract reactances
Z = √(R² + (XL - XC)²)
Impedance Z = √(R² + (XL - XC)²). Calculate XL = ωL and XC = 1/ωC separately.
XL = 1000×0.9 = 900Ω, XC = 1/(1000×2×10⁻⁶) = 500Ω
Z = √(300² + (900-500)²) = √(90000 + 160000) = 500Ω
Adding instead of subtracting reactances
Series resonant circuit, C=10⁻⁶F, L=10⁻⁴H. Frequency of oscillations:
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f = 1/(2π√LC)
Resonance frequency formula is f = 1/(2π√LC). Let's substitute our values carefully.
f = 1/(2π√(10⁻⁴×10⁻⁶)) = 1/(2π√10⁻¹⁰) = 10⁵/2π Hz
Always check units: L in henry, C in farad gives frequency in hertz!
Unit conversion errors
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Think about LC oscillation formula
f = 1/(2π√LC)
Resonance frequency formula is f = 1/(2π√LC). Let's substitute our values carefully.
f = 1/(2π√(10⁻⁴×10⁻⁶)) = 1/(2π√10⁻¹⁰) = 10⁵/2π Hz
Always check units: L in henry, C in farad gives frequency in hertz!
Unit conversion errors
Charged capacitor C, initial charge q₀ connected to inductor L at t=0. Time for equal electric and magnetic energy is:
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T = 2π√LC
Initially, all energy is electric (½q²/C). As oscillation starts, energy converts between electric and magnetic forms.
Equal energy means ½q²/C = ½LI². This occurs at ¼ period of oscillation.
Period T = 2π√LC, so time for equal energy = T/4 = π√LC/4
Forgetting energy conversion
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Consider when energies exchange
T = 2π√LC
Initially, all energy is electric (½q²/C). As oscillation starts, energy converts between electric and magnetic forms.
Equal energy means ½q²/C = ½LI². This occurs at ¼ period of oscillation.
Period T = 2π√LC, so time for equal energy = T/4 = π√LC/4
Forgetting energy conversion
Alternating current i = i₁sinωt + i₂cosωt. RMS current is:
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I_rms = √(average of i²)
RMS value is root mean square over one cycle. Need to integrate (i₁sinωt + i₂cosωt)² over period.
Sin and cos terms are orthogonal - their product averages to zero over complete cycle.
Therefore, I_rms = √[(i₁²/2 + i₂²/2)] = √(i₁² + i₂²)/2
Forgetting orthogonality
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Remember sin and cos are orthogonal
I_rms = √(average of i²)
RMS value is root mean square over one cycle. Need to integrate (i₁sinωt + i₂cosωt)² over period.
Sin and cos terms are orthogonal - their product averages to zero over complete cycle.
Therefore, I_rms = √[(i₁²/2 + i₂²/2)] = √(i₁² + i₂²)/2
Forgetting orthogonality
Wire stretched 10%. New resistance and specific resistance becomes:
🥳 Wohoo! Correct answer
R = ρl/A
When wire is stretched, length increases by 10% (l₂ = 1.1l₁), and area decreases (A₂ = A₁/1.1) due to constant volume.
Resistance R = ρl/A. New resistance = ρ(1.1l)/((A/1.1)) = 1.21R₁
Specific resistance (ρ) is material property, doesn't change with shape!
Thinking ρ changes
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Think about volume conservation
R = ρl/A
When wire is stretched, length increases by 10% (l₂ = 1.1l₁), and area decreases (A₂ = A₁/1.1) due to constant volume.
Resistance R = ρl/A. New resistance = ρ(1.1l)/((A/1.1)) = 1.21R₁
Specific resistance (ρ) is material property, doesn't change with shape!
Thinking ρ changes
If voltage across 220V, 100W bulb drops 2.5%, power decrease percentage is:
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P = V²/R
Power is proportional to voltage squared (P = V²/R). When voltage changes, power changes quadratically.
If V₂ = 0.975V₁ (2.5% decrease), then P₂/P₁ = (V₂/V₁)² = (0.975)²
Therefore, P₂/P₁ = 0.95, meaning power decreases by 5%. Notice it's double the voltage percentage!
Linear vs quadratic change
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Remember P ∝ V²
P = V²/R
Power is proportional to voltage squared (P = V²/R). When voltage changes, power changes quadratically.
If V₂ = 0.975V₁ (2.5% decrease), then P₂/P₁ = (V₂/V₁)² = (0.975)²
Therefore, P₂/P₁ = 0.95, meaning power decreases by 5%. Notice it's double the voltage percentage!
Linear vs quadratic change
Galvanometer resistance 50Ω, 3V battery, 2950Ω series resistance gives 30 divisions. For 20 divisions, series resistance should be:
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I = V/R
Current and deflection are proportional in galvanometer. If I₁/I₂ = θ₁/θ₂, then (30/20) = (R₂/R₁), where R is total resistance.
Original total R₁ = 3000Ω (2950 + 50). For 20 divisions, new total R₂ = R₁×(30/20) = 4500Ω
Therefore, new series resistance = 4500 - 50 = 4450Ω (subtract galvanometer resistance)
Forgetting galvanometer resistance
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Use current ratio equals deflection ratio
I = V/R
Current and deflection are proportional in galvanometer. If I₁/I₂ = θ₁/θ₂, then (30/20) = (R₂/R₁), where R is total resistance.
Original total R₁ = 3000Ω (2950 + 50). For 20 divisions, new total R₂ = R₁×(30/20) = 4500Ω
Therefore, new series resistance = 4500 - 50 = 4450Ω (subtract galvanometer resistance)
Forgetting galvanometer resistance
Ten identical cells each of potential E and internal resistance r in series. Voltmeter across three cells reads:
🥳 Wohoo! Correct answer
V = N×E
In a series combination, total EMF is sum of individual EMFs. With 10 identical cells, total EMF is 10E.
When we measure across 3 cells, we're only measuring the EMF of those 3 cells.
Therefore, voltmeter reading = 3E. Internal resistance doesn't affect this reading since voltmeter has very high resistance!
Ignoring cell position
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Think about how voltages add in series
V = N×E
In a series combination, total EMF is sum of individual EMFs. With 10 identical cells, total EMF is 10E.
When we measure across 3 cells, we're only measuring the EMF of those 3 cells.
Therefore, voltmeter reading = 3E. Internal resistance doesn't affect this reading since voltmeter has very high resistance!
Ignoring cell position
When metal conductor in left gap of meter bridge is heated, balancing point:
🥳 Wohoo! Correct answer
R = R₀(1 + αΔT)
When we heat a metal, its resistance increases. In meter bridge, we're comparing unknown resistance with known resistance.
The balancing equation is R/S = L/(100-L), where L is length. If R increases, L must increase to maintain balance.
Therefore, balance point shifts right. Think about it: we need more wire length on left to balance increased resistance!
Forgetting temperature effect
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Remember resistance increases with temperature
R = R₀(1 + αΔT)
When we heat a metal, its resistance increases. In meter bridge, we're comparing unknown resistance with known resistance.
The balancing equation is R/S = L/(100-L), where L is length. If R increases, L must increase to maintain balance.
Therefore, balance point shifts right. Think about it: we need more wire length on left to balance increased resistance!
Forgetting temperature effect
Electron in atom revolves radius 0.72Å, making 9.4×10¹⁸ revolutions/s. Current is [e=1.6×10⁻¹⁹C]:
🥳 Wohoo! Correct answer
I = ef
Current is charge flowing per second. Here, one electron completes 9.4×10¹⁸ revolutions/s. Each revolution carries charge e=1.6×10⁻¹⁹C.
Current = charge/time = (charge per revolution × revolutions per second)
I = (1.6×10⁻¹⁹)(9.4×10¹⁸) = 1.5A. The units work: C/s = A
Forgetting frequency factor
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Consider electron as tiny current loop
I = ef
Current is charge flowing per second. Here, one electron completes 9.4×10¹⁸ revolutions/s. Each revolution carries charge e=1.6×10⁻¹⁹C.
Current = charge/time = (charge per revolution × revolutions per second)
I = (1.6×10⁻¹⁹)(9.4×10¹⁸) = 1.5A. The units work: C/s = A
Forgetting frequency factor
Wire-bound resistors are made by:
🥳 Wohoo! Correct answer
R = ρl/A
Think about what properties we need in a resistor: stable resistance, low temperature coefficient, and durability. Alloys like Manganin and Constantan were specifically developed for this!
Pure metals like Cu, Al, or Ag have low resistance and high temperature coefficients - not good for resistors. Semiconductor materials like Ge aren't used for wire-wound resistors.
Manganin (Cu-Mn-Ni alloy), Constantan (Cu-Ni alloy), and Nichrome are perfect because they have high resistivity, low temperature coefficient, and don't oxidize easily.
Thinking any metal works
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Consider why we need special alloys for resistors
R = ρl/A
Think about what properties we need in a resistor: stable resistance, low temperature coefficient, and durability. Alloys like Manganin and Constantan were specifically developed for this!
Pure metals like Cu, Al, or Ag have low resistance and high temperature coefficients - not good for resistors. Semiconductor materials like Ge aren't used for wire-wound resistors.
Manganin (Cu-Mn-Ni alloy), Constantan (Cu-Ni alloy), and Nichrome are perfect because they have high resistivity, low temperature coefficient, and don't oxidize easily.
Thinking any metal works
Charged particle in E=3×10¹⁰ V/m⁻¹, mobility 2.5×10⁻⁶ m²/V/s, drift velocity is:
🥳 Wohoo! Correct answer
vd = μE
Drift velocity is related to mobility and electric field strength. The formula is vd = μE, where μ is mobility. Let's identify our values: μ = 2.5×10⁻⁶ m²/V/s, E = 3×10¹⁰ V/m
Simply multiply these numbers: vd = (2.5×10⁻⁶)(3×10¹⁰). Keep track of units: (m²/V/s)(V/m) = m/s
This gives us 7.5×10⁴ m/s. This is reasonable for drift velocity in strong fields!
Confusing units
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Think about how mobility relates field to velocity
vd = μE
Drift velocity is related to mobility and electric field strength. The formula is vd = μE, where μ is mobility. Let's identify our values: μ = 2.5×10⁻⁶ m²/V/s, E = 3×10¹⁰ V/m
Simply multiply these numbers: vd = (2.5×10⁻⁶)(3×10¹⁰). Keep track of units: (m²/V/s)(V/m) = m/s
This gives us 7.5×10⁴ m/s. This is reasonable for drift velocity in strong fields!
Confusing units
In series LCR circuit at resonance, the phase difference between voltage and current is
🥳 Wohoo! Correct answer
φ = tan⁻¹(XL-XC)/R
At resonance: XL=XC
Net reactive impedance = 0
Phase angle = tan⁻¹(0) = 0°
Impedance confusion
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Consider impedance triangle
φ = tan⁻¹(XL-XC)/R
At resonance: XL=XC
Net reactive impedance = 0
Phase angle = tan⁻¹(0) = 0°
Impedance confusion
A moving coil galvanometer is converted into an ammeter of range 0 to 5 mA. The galvanometer resistance is 90Ω and the shunt resistance has a value of 10Ω. If there are 50 divisions in the galvanometer turned ammeter on either sides of zero, its current sensitivity is
🥳 Wohoo! Correct answer
S = N/I where N=divisions, I=current
Calculate current per division: Total current range = 5mA = 0.005A, Number of divisions = 50
Find division per ampere: 50 div/0.005A = 10,000 div/A = 1×10⁵ div/A
Verify units match option (div/A)
Confusing A/div with div/A
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Remember sensitivity is divisions per unit current
S = N/I where N=divisions, I=current
Calculate current per division: Total current range = 5mA = 0.005A, Number of divisions = 50
Find division per ampere: 50 div/0.005A = 10,000 div/A = 1×10⁵ div/A
Verify units match option (div/A)
Confusing A/div with div/A
For a given electric current the drift velocity of conduction electrons in a copper wire is vd and their mobility is μ. When the current is increased at constant temperature
🥳 Wohoo! Correct answer
vd = μE
Current I ∝ drift velocity vd
Mobility μ depends only on temperature
At constant T, μ stays same but vd increases
Mobility temperature dependence
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Temperature kept constant
vd = μE
Current I ∝ drift velocity vd
Mobility μ depends only on temperature
At constant T, μ stays same but vd increases
Mobility temperature dependence
A wire of resistance R is connected across a cell of emf ε and internal resistance r. The current through the circuit is I. In time t, the work done by battery to establish the current I is
🥳 Wohoo! Correct answer
W = VIt = I²Rt
Work = Power × Time
Power in resistor = I²R
W = I²R × t
Mixing power formulas
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Consider both V and I
W = VIt = I²Rt
Work = Power × Time
Power in resistor = I²R
W = I²R × t
Mixing power formulas
Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?
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Q=(1/R)√(L/C)
Calculate Q-factor using Q=(1/R)√(L/C)
Compare Q-factors for all combinations
Highest Q-factor gives best tuning
Q-factor calculation
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Consider quality factor
Q=(1/R)√(L/C)
Calculate Q-factor using Q=(1/R)√(L/C)
Compare Q-factors for all combinations
Highest Q-factor gives best tuning
Q-factor calculation
E is the electric field inside a conductor whose material has conductivity σ and resistivity ρ. The current density inside the conductor is j. The correct form of Ohm's law is
🥳 Wohoo! Correct answer
E=ρj, j=σE
Start with basic Ohm's law V=IR. Convert to point form by considering voltage per unit length and current per unit area
Express E=V/l and j=I/A where E is electric field and j is current density
Substitute to get E=ρj, where ρ is resistivity
Confusing σ and ρ
😢 Uh oh! Incorrect answer, Try again
Think about units of each quantity
E=ρj, j=σE
Start with basic Ohm's law V=IR. Convert to point form by considering voltage per unit length and current per unit area
Express E=V/l and j=I/A where E is electric field and j is current density
Substitute to get E=ρj, where ρ is resistivity
Confusing σ and ρ
Ten identical cells each emf 2V and internal resistances 1Ω are connected in series with two cells wrongly connected. A resistor of 10Ω is connected to the combination. What is the current through the resistor?
🥳 Wohoo! Correct answer
I = (n-2m)ε/(R+nr)
8 cells forward (16V) + 2 cells reversed (-4V) gives net EMF = 12V
Total resistance = 10 cells × 1Ω + external 10Ω = 20Ω
I = 12V/20Ω = 0.6A
Forgetting internal resistance
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Reversed cells reduce EMF but add resistance
I = (n-2m)ε/(R+nr)
8 cells forward (16V) + 2 cells reversed (-4V) gives net EMF = 12V
Total resistance = 10 cells × 1Ω + external 10Ω = 20Ω
I = 12V/20Ω = 0.6A
Forgetting internal resistance
An electric bulb of 60W, 120V is to be connected to 220V source. What resistance should be connected in series with the bulb, so that the bulb glows properly?
🥳 Wohoo! Correct answer
P=V²/R
Calculate bulb resistance: R=V²/P=(120)²/60=240Ω
Find current through bulb: I=P/V=60/120=0.5A
Required series resistance=(220-120)/0.5=200Ω
Power calculation errors
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Consider power rating
P=V²/R
Calculate bulb resistance: R=V²/P=(120)²/60=240Ω
Find current through bulb: I=P/V=60/120=0.5A
Required series resistance=(220-120)/0.5=200Ω
Power calculation errors
The resistance of a carbon resistor is 4.7kΩ±5%. The colour of the third band is
🥳 Wohoo! Correct answer
R = digit-digit-multiplier
4.7kΩ = 47 × 102Ω
Third band represents multiplier (×102)
Red band indicates multiplier of 100
Wrong multiplier color
😢 Uh oh! Incorrect answer, Try again
Remember color code sequence
R = digit-digit-multiplier
4.7kΩ = 47 × 102Ω
Third band represents multiplier (×102)
Red band indicates multiplier of 100
Wrong multiplier color
The four bands of a colour coded resistor are of the colour grey, red, gold and gold. The value of the resistance of the resistor is
🥳 Wohoo! Correct answer
R = 82×0.1Ω±5%
Grey(8), Red(2) give 82
First gold = ×0.1 multiplier
Second gold = ±5% tolerance
Confusing gold bands
😢 Uh oh! Incorrect answer, Try again
Note both golds' meanings
R = 82×0.1Ω±5%
Grey(8), Red(2) give 82
First gold = ×0.1 multiplier
Second gold = ±5% tolerance
Confusing gold bands