Current Electricity
Q. If a resistor has a resistance of 5 ohms and a current of 2 amperes flows through it, what is the voltage across the resistor?
A.
10 V
B.
5 V
C.
2.5 V
D.
1 V
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Solution
Using Ohm's Law, V = I * R = 2 A * 5 Ω = 10 V.
Correct Answer: A — 10 V
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Q. If a resistor is connected to a 9V battery and the current flowing through it is 3A, what is the resistance of the resistor?
A.
1 Ω
B.
3 Ω
C.
9 Ω
D.
27 Ω
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Solution
Using Ohm's Law, R = V / I = 9 V / 3 A = 3 Ω.
Correct Answer: C — 9 Ω
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Q. If a wire's length is doubled while keeping its cross-sectional area constant, how does its resistance change?
A.
Remains the same
B.
Doubles
C.
Halves
D.
Quadruples
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Solution
Resistance is directly proportional to length; doubling the length doubles the resistance.
Correct Answer: B — Doubles
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Q. If one of the resistors in a Wheatstone bridge is replaced with a variable resistor, what is the effect on the balance condition?
A.
It cannot be balanced
B.
It can be balanced by adjusting the variable resistor
C.
It will always be unbalanced
D.
It will short-circuit the bridge
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Solution
The variable resistor allows for adjustment to achieve balance in the bridge.
Correct Answer: B — It can be balanced by adjusting the variable resistor
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Q. If R1 = 10Ω, R2 = 15Ω, R3 = 5Ω, what should R4 be for the Wheatstone bridge to be balanced?
A.
7.5Ω
B.
10Ω
C.
12.5Ω
D.
15Ω
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Solution
Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (15 * 5) / 10 = 7.5Ω.
Correct Answer: C — 12.5Ω
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Q. If R1 = 10Ω, R2 = 20Ω, and R3 = 30Ω in a Wheatstone bridge, what should R4 be for the bridge to be balanced?
A.
15Ω
B.
20Ω
C.
25Ω
D.
30Ω
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Solution
For balance, R1/R2 = R3/R4, thus R4 = (R2 * R3) / R1 = (20 * 30) / 10 = 60Ω.
Correct Answer: B — 20Ω
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Q. If R1 = 10Ω, R2 = 20Ω, R3 = 30Ω, and R4 = 60Ω in a Wheatstone bridge, is the bridge balanced?
A.
Yes
B.
No
C.
Depends on the voltage
D.
Depends on the current
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Solution
The bridge is not balanced because 10/20 ≠ 30/60.
Correct Answer: B — No
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Q. If R1 = 2Ω, R2 = 3Ω, and R3 = 6Ω in a Wheatstone bridge, what is the value of R4 for balance?
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Solution
Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.
Correct Answer: A — 4Ω
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Q. If R1 = 2Ω, R2 = 3Ω, and R3 = 6Ω, what is the value of R4 for the Wheatstone bridge to be balanced?
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Solution
Using the balance condition R1/R2 = R3/R4, we have 2/3 = 6/R4. Solving gives R4 = 4Ω.
Correct Answer: A — 4Ω
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Q. If R1 = 5Ω, R2 = 10Ω, and R3 = 15Ω in a Wheatstone bridge, what is the value of R4 for balance?
A.
7.5Ω
B.
10Ω
C.
15Ω
D.
20Ω
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Solution
Using the balance condition R1/R2 = R3/R4, we find R4 = (R2 * R3) / R1 = (10 * 15) / 5 = 30Ω.
Correct Answer: B — 10Ω
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Q. If R1 = 5Ω, R2 = 15Ω, R3 = 10Ω, what should R4 be for the Wheatstone bridge to be balanced?
A.
30Ω
B.
10Ω
C.
15Ω
D.
5Ω
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Solution
For balance, R4 must be 30Ω since 5/15 = 10/30.
Correct Answer: B — 10Ω
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Q. If the balance length of a potentiometer is 50cm for a cell of unknown emf, and the potential gradient is 4 V/m, what is the emf of the cell?
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Solution
The emf is calculated as V = potential gradient × length = 4 V/m × 0.5 m = 2 V.
Correct Answer: B — 4V
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Q. If the balancing length of a potentiometer is found to be 40 cm for a cell of emf 2V, what is the potential gradient if the total length of the wire is 100 cm?
A.
5 V/m
B.
2 V/m
C.
4 V/m
D.
3 V/m
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Solution
The potential gradient is V/L = 2V/0.4m = 5 V/m.
Correct Answer: A — 5 V/m
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Q. If the cross-sectional area of a wire is doubled, how does its resistance change?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Increases four times
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Solution
Resistance is inversely proportional to cross-sectional area; doubling the area halves the resistance.
Correct Answer: B — Halves
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Q. If the cross-sectional area of a wire is doubled, what happens to its resistance?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Increases four times
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Solution
Resistance (R) is inversely proportional to cross-sectional area (A). Doubling A halves the resistance.
Correct Answer: B — Halves
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Q. If the current in a circuit is 5 A and the resistance is 2 ohms, what is the power consumed in the circuit?
A.
10 W
B.
25 W
C.
5 W
D.
2.5 W
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Solution
Power (P) can be calculated using P = I^2R = (5 A)^2 * 2 Ω = 25 W.
Correct Answer: B — 25 W
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Q. If the current in a circuit is doubled while the resistance remains constant, what happens to the voltage?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to Ohm's Law, if the current (I) is doubled and resistance (R) remains constant, the voltage (V) must also double.
Correct Answer: A — It doubles
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Q. If the current in a circuit is halved, what happens to the power consumed by the circuit?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Power P = I^2 * R. If I is halved, P becomes (1/2)^2 * R = 1/4 * P, so the power is reduced to a quarter, which means it halves.
Correct Answer: B — It halves
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Q. If the current in a circuit is split into two branches with resistances R1 = 2Ω and R2 = 4Ω, what is the current through R1 if the total current is 6A?
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Solution
Using the current division rule, I1 = I_total * (R2 / (R1 + R2)) = 6A * (4Ω / (2Ω + 4Ω)) = 6A * (4/6) = 4A.
Correct Answer: B — 3A
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Q. If the current through a 10Ω resistor is 2A, what is the voltage across the resistor?
A.
5V
B.
10V
C.
20V
D.
30V
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Solution
Using Ohm's law, V = I * R = 2A * 10Ω = 20V.
Correct Answer: C — 20V
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Q. If the current through a conductor is doubled while the resistance remains constant, what happens to the power consumed?
A.
It doubles
B.
It triples
C.
It remains the same
D.
It halves
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Solution
Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is four times the original power.
Correct Answer: A — It doubles
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Q. If the current through a conductor is doubled, what happens to the power dissipated in the conductor?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is quadrupled.
Correct Answer: B — It quadruples
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Q. If the current through a conductor is doubled, what happens to the resistance if the voltage remains constant?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to Ohm's law, V = I * R. If V is constant and I is doubled, R must halve to maintain the equation.
Correct Answer: B — It halves
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Q. If the current through a resistor is 3A and the resistance is 4Ω, what is the voltage across the resistor?
A.
6V
B.
9V
C.
12V
D.
15V
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Solution
Using Ohm's law, V = I * R = 3A * 4Ω = 12V.
Correct Answer: C — 12V
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Q. If the current through a resistor is doubled, what happens to the power consumed by the resistor?
A.
Increases by 2 times
B.
Increases by 4 times
C.
Remains the same
D.
Decreases by 2 times
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Solution
Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is 4 times the original power.
Correct Answer: B — Increases by 4 times
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Q. If the current through a resistor is doubled, what happens to the power dissipated in the resistor?
A.
It halves
B.
It doubles
C.
It quadruples
D.
It remains the same
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Solution
Power P = I^2 * R. If I is doubled, P becomes (2I)^2 * R = 4I^2 * R, which is quadrupled.
Correct Answer: C — It quadruples
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Q. If the emf of a cell is 12 V and the potentiometer wire is 20 m long, what is the potential gradient if the wire is calibrated to give a reading of 0.6 V/m?
A.
12 V
B.
0.6 V/m
C.
0.3 V/m
D.
0.5 V/m
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Solution
The potential gradient is given as 0.6 V/m, which is the calibration value for the potentiometer.
Correct Answer: B — 0.6 V/m
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Q. If the emf of a cell is 2V and the potential gradient of the potentiometer wire is 4 V/m, what will be the balancing length?
A.
0.5 m
B.
1 m
C.
0.25 m
D.
0.75 m
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Solution
The balancing length is calculated as L = V / (potential gradient) = 2V / 4 V/m = 0.5 m.
Correct Answer: B — 1 m
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Q. If the galvanometer in a Wheatstone bridge has a high resistance, what is the effect on the sensitivity of the bridge?
A.
Increases sensitivity
B.
Decreases sensitivity
C.
No effect
D.
Depends on the resistances
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Solution
A high resistance in the galvanometer increases the sensitivity of the bridge.
Correct Answer: A — Increases sensitivity
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Q. If the galvanometer in a Wheatstone bridge has a resistance of 5Ω and the bridge is balanced, what is the current through the galvanometer?
A.
Zero.
B.
5A.
C.
1A.
D.
10A.
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Solution
In a balanced Wheatstone bridge, the current through the galvanometer is zero.
Correct Answer: A — Zero.
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