Circuits & Kirchhoff Laws
Q. If a circuit has a total resistance of 10Ω and a current of 2A flowing through it, what is the total voltage supplied by the battery?
A.
5V
B.
10V
C.
15V
D.
20V
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Solution
Using Ohm's Law, V = I * R = 2A * 10Ω = 20V.
Correct Answer: B — 10V
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Q. If a circuit has a total resistance of 10Ω and a current of 2A, what is the total voltage supplied by the battery?
A.
10V
B.
20V
C.
30V
D.
40V
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Solution
Using Ohm's Law, V = I * R = 2A * 10Ω = 20V.
Correct Answer: B — 20V
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Q. If a circuit has a total resistance of 10Ω and a current of 3A, what is the total voltage supplied by the battery?
A.
10V
B.
20V
C.
30V
D.
40V
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Solution
Using Ohm's Law, V = I * R = 3A * 10Ω = 30V.
Correct Answer: C — 30V
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Q. If a circuit has a total resistance of 10Ω and a current of 5A, what is the total voltage supplied by the battery?
A.
10V
B.
20V
C.
30V
D.
50V
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Solution
Using Ohm's Law, V = I * R = 5A * 10Ω = 50V.
Correct Answer: B — 20V
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Q. If a circuit has a total resistance of 12Ω and a current of 1.5A flowing through it, what is the total voltage supplied by the battery?
A.
18V
B.
12V
C.
6V
D.
9V
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Solution
Using Ohm's Law, V = I * R = 1.5A * 12Ω = 18V.
Correct Answer: A — 18V
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Q. If a circuit has a total resistance of 12Ω and a current of 1.5A flows through it, what is the total voltage supplied by the battery?
A.
18V
B.
12V
C.
6V
D.
24V
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Solution
Using Ohm's Law, V = I * R = 1.5A * 12Ω = 18V.
Correct Answer: A — 18V
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Q. If a circuit has a total resistance of 12Ω and a current of 3A, what is the voltage supplied by the battery?
A.
36V
B.
24V
C.
12V
D.
9V
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Solution
Using Ohm's Law, V = I * R = 3A * 12Ω = 36V.
Correct Answer: A — 36V
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Q. If a circuit has a total voltage of 10V and a total current of 2A, what is the total resistance in the circuit?
A.
5Ω
B.
10Ω
C.
15Ω
D.
20Ω
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Solution
Using Ohm's Law (R = V/I), total resistance R = 10V/2A = 5Ω.
Correct Answer: A — 5Ω
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Q. If a circuit has a total voltage of 24V and two resistors of 6Ω and 3Ω in series, what is the current flowing through the circuit?
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Solution
Total resistance R_total = 6Ω + 3Ω = 9Ω. Current I = V/R = 24V / 9Ω = 2.67A.
Correct Answer: B — 3A
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Q. If a circuit has a total voltage of 24V and two resistors of 8Ω and 4Ω in series, what is the current flowing through the circuit?
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Solution
Total resistance R_total = 8Ω + 4Ω = 12Ω. Using Ohm's law, I = V/R = 24V/12Ω = 2A.
Correct Answer: B — 2A
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Q. If a circuit has two branches with resistances of 4Ω and 8Ω, what is the total current if the voltage across the branches is 12V?
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Solution
For the 4Ω branch, I1 = V/R1 = 12V / 4Ω = 3A. For the 8Ω branch, I2 = V/R2 = 12V / 8Ω = 1.5A. Total current = I1 + I2 = 3A + 1.5A = 4.5A.
Correct Answer: B — 2A
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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 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 voltage across a resistor is doubled, what happens to the current through the resistor, assuming resistance remains constant?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Increases by a factor of four
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Solution
According to Ohm's Law (V = IR), if voltage is doubled and resistance remains constant, current also doubles.
Correct Answer: A — Doubles
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Q. If two resistors of 4 ohms and 6 ohms are connected in parallel, what is the current through the 4 ohm resistor when a 12V battery is connected?
A.
3 A
B.
2 A
C.
1 A
D.
4 A
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Solution
Using Ohm's law, I = V/R, for the 4 ohm resistor, I = 12V / 4 ohms = 3 A.
Correct Answer: A — 3 A
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Q. If two resistors of 4 ohms and 6 ohms are connected in series, what is the voltage drop across the 4 ohm resistor when a current of 2 A flows through the circuit?
A.
8 V
B.
4 V
C.
2 V
D.
6 V
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Solution
Using Ohm's law, V = I * R, the voltage drop across the 4 ohm resistor is V = 2 A * 4 ohms = 8 V.
Correct Answer: A — 8 V
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Q. In a circuit with a 10V battery and two resistors (3Ω and 6Ω) in series, what is the voltage drop across the 6Ω resistor?
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Solution
Voltage drop across R2 = (R2 / (R1 + R2)) * Vtotal = (6 / (3 + 6)) * 10 = 6.67V.
Correct Answer: C — 6V
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Q. In a circuit with a 10V battery and two resistors in parallel (4Ω and 6Ω), what is the voltage across each resistor?
A.
10V for both
B.
5V for both
C.
10V for 4Ω and 6V for 6Ω
D.
10V for 6Ω and 4V for 4Ω
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Solution
In a parallel circuit, the voltage across each resistor is the same as the source voltage. Therefore, both resistors have 10V across them.
Correct Answer: A — 10V for both
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Q. In a circuit with a 10V battery and two resistors in series (2Ω and 3Ω), what is the voltage drop across the 3Ω resistor?
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Solution
Using the voltage divider rule, V3 = (R3 / (R2 + R3)) * Vtotal = (3 / (2 + 3)) * 10 = 6V.
Correct Answer: B — 6V
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Q. In a circuit with a 10V battery and two resistors in series (R1 = 2Ω, R2 = 3Ω), what is the current flowing through the circuit?
A.
2A
B.
1A
C.
0.5A
D.
3A
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Solution
Total resistance R_total = R1 + R2 = 2Ω + 3Ω = 5Ω. Current I = V / R_total = 10V / 5Ω = 2A.
Correct Answer: B — 1A
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Q. In a circuit with a 10V battery and two resistors in series (R1 = 2Ω, R2 = 8Ω), what is the current flowing through the circuit?
A.
0.5A
B.
1A
C.
2A
D.
5A
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Solution
Total resistance R_total = R1 + R2 = 2Ω + 8Ω = 10Ω. Current I = V/R = 10V/10Ω = 1A.
Correct Answer: B — 1A
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Q. In a circuit with a 10V battery and two resistors of 5Ω each in series, what is the voltage across the second resistor?
A.
5V
B.
10V
C.
2.5V
D.
0V
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Solution
In series, the voltage is divided equally: V2 = (R2 / (R1 + R2)) * Vtotal = (5 / (5 + 5)) * 10 = 5V.
Correct Answer: A — 5V
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Q. In a circuit with a 12 V battery and two resistors in series (4 ohms and 8 ohms), what is the voltage drop across the 8 ohm resistor?
A.
4 V
B.
8 V
C.
6 V
D.
12 V
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Solution
The total resistance is 4 + 8 = 12 ohms. The current I = V / R = 12 V / 12 ohms = 1 A. The voltage drop across the 8 ohm resistor is V = I * R = 1 A * 8 ohms = 8 V.
Correct Answer: B — 8 V
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Q. In a circuit with a 12V battery and two resistors in parallel (4 ohms and 6 ohms), what is the total current supplied by the battery?
A.
3 A
B.
2 A
C.
1 A
D.
4 A
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Solution
First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 = 5/12, so R_eq = 12/5 = 2.4 ohms. Then, using Ohm's law, I = V/R, we have I = 12V / 2.4 ohms = 5 A.
Correct Answer: A — 3 A
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Q. In a circuit with a 12V battery and two resistors in series (3 ohms and 6 ohms), what is the voltage drop across the 6 ohm resistor?
A.
8 V
B.
4 V
C.
6 V
D.
12 V
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Solution
The total resistance is 9 ohms. The current is I = V/R = 12V / 9 ohms = 4/3 A. The voltage drop across the 6 ohm resistor is V = I * R = (4/3 A) * 6 ohms = 8 V.
Correct Answer: A — 8 V
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Q. In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the current flowing through the circuit?
A.
0.5A
B.
1A
C.
1.5A
D.
2A
Show solution
Solution
Total resistance R = R1 + R2 = 4Ω + 8Ω = 12Ω. Using Ohm's law, I = V/R = 12V / 12Ω = 1A.
Correct Answer: B — 1A
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Q. In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω), what is the current flowing through the circuit?
A.
0.5A
B.
1A
C.
1.5A
D.
2A
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Solution
Total resistance R_total = R1 + R2 = 4Ω + 8Ω = 12Ω. Current I = V/R_total = 12V/12Ω = 1A.
Correct Answer: B — 1A
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Q. In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω), what is the voltage across R2?
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Solution
The total resistance is R_total = R1 + R2 = 4Ω + 8Ω = 12Ω. The current is I = V/R_total = 12V / 12Ω = 1A. The voltage across R2 is V_R2 = I * R2 = 1A * 8Ω = 8V.
Correct Answer: C — 8V
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