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Q. In a circuit with a 24V battery and three resistors in parallel (4Ω, 6Ω, and 12Ω), what is the total current supplied by the battery?

A. 2A
B. 3A
C. 4A
D. 6A

Solution

1/R_eq = 1/4 + 1/6 + 1/12 = 1. Therefore, R_eq = 2Ω. Total current I = V/R_eq = 24V/2Ω = 12A.

Correct Answer: C — 4A

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Q. In a circuit with a 24V battery and three resistors in series (R1 = 2Ω, R2 = 3Ω, R3 = 5Ω), what is the voltage drop across R2?

A. 6V
B. 8V
C. 12V
D. 10V

Solution

Total resistance R_total = R1 + R2 + R3 = 2Ω + 3Ω + 5Ω = 10Ω. Current I = V/R_total = 24V / 10Ω = 2.4A. Voltage drop across R2 = I * R2 = 2.4A * 3Ω = 7.2V.

Correct Answer: B — 8V

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Q. In a circuit with a 24V battery and three resistors of 2 ohms, 4 ohms, and 6 ohms in series, what is the total current flowing through the circuit?

A. 2 A
B. 3 A
C. 4 A
D. 1 A

Solution

Total resistance R = 2 + 4 + 6 = 12 ohms. Using Ohm's law, I = V/R = 24V / 12 ohms = 2 A.

Correct Answer: B — 3 A

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Q. In a circuit with a 24V battery and three resistors of 4Ω, 4Ω, and 8Ω in series, what is the voltage drop across the 8Ω resistor?

A. 8V
B. 12V
C. 16V
D. 24V

Solution

Total resistance R = 4Ω + 4Ω + 8Ω = 16Ω. Current I = V/R = 24V/16Ω = 1.5A. Voltage drop across 8Ω = I * R = 1.5A * 8Ω = 12V.

Correct Answer: B — 12V

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Q. In a circuit with a 24V battery and two 12 ohm resistors in series, what is the current flowing through the circuit?

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Solution

Total resistance R_total = 12 + 12 = 24 ohms. Current I = V/R_total = 24V / 24Ω = 1 A.

Correct Answer: B — 2 A

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Q. In a circuit with a 24V battery and two 12Ω resistors in parallel, what is the total current supplied by the battery?

A. 1A
B. 2A
C. 3A
D. 4A

Solution

Total resistance R = 1/(1/12 + 1/12) = 6Ω; Total current I = V/R = 24V / 6Ω = 4A.

Correct Answer: D — 4A

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Q. In a circuit with a 24V battery and two 12Ω resistors in series, what is the current flowing through the circuit?

A. 1A
B. 2A
C. 3A
D. 4A

Solution

Total resistance R_total = 12Ω + 12Ω = 24Ω. Current I = V/R_total = 24V / 24Ω = 1A.

Correct Answer: B — 2A

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Q. In a circuit with a 24V battery and two resistors in series (4Ω and 8Ω), what is the voltage across the 8Ω resistor?

A. 8V
B. 12V
C. 16V
D. 20V

Solution

Voltage across 8Ω resistor = (R2 / (R1 + R2)) * Vtotal = (8 / (4 + 8)) * 24 = 16V.

Correct Answer: B — 12V

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Q. In a circuit with a 24V battery and two resistors in series (6Ω and 3Ω), what is the voltage across the 3Ω resistor?

A. 8V
B. 12V
C. 16V
D. 4V

Solution

Total resistance = 6Ω + 3Ω = 9Ω. Current I = V / R = 24V / 9Ω = 2.67A. Voltage across 3Ω = I * R = 2.67A * 3Ω = 8V.

Correct Answer: A — 8V

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Q. In a circuit with a 24V battery and two resistors in series (R1 = 6Ω, R2 = 12Ω), what is the voltage drop across R2?

A. 12V
B. 18V
C. 6V
D. 8V

Solution

Voltage drop across R2 = V_R2 = (R2 / (R1 + R2)) * V_total = (12Ω / (6Ω + 12Ω)) * 24V = 12V.

Correct Answer: A — 12V

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Q. In a circuit with a 24V battery and two resistors in series (R1 = 8Ω, R2 = 16Ω), what is the voltage across R2?

A. 16V
B. 8V
C. 12V
D. 4V

Solution

Using voltage division, V_R2 = V_total * (R2 / (R1 + R2)) = 24V * (16Ω / (8Ω + 16Ω)) = 16V.

Correct Answer: A — 16V

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Q. In a circuit with a 24V battery and two resistors in series (R1 = 8Ω, R2 = 16Ω), what is the current flowing through the circuit?

A. 1A
B. 2A
C. 3A
D. 4A

Solution

Total resistance R_total = R1 + R2 = 8Ω + 16Ω = 24Ω. The current is I = V/R_total = 24V / 24Ω = 1A.

Correct Answer: B — 2A

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Q. In a circuit with a 24V battery and two resistors of 4Ω and 8Ω in series, what is the voltage across the 8Ω resistor?

A. 8V
B. 12V
C. 16V
D. 20V

Solution

Voltage across R2 = (R2 / (R1 + R2)) * V = (8 / (4 + 8)) * 24 = 16V.

Correct Answer: B — 12V

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Q. In a circuit with a 24V battery and two resistors of 6Ω and 12Ω in series, what is the voltage across the 12Ω resistor?

A. 8V
B. 12V
C. 16V
D. 6V

Solution

Voltage across R2 = (R2/(R1+R2)) * V = (12/(6+12)) * 24 = 12V.

Correct Answer: B — 12V

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Q. In a circuit with a 24V battery and two resistors of 8 ohms and 16 ohms in series, what is the total current in the circuit?

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Solution

Total resistance R_total = 8 + 16 = 24 ohms. Current I = V/R_total = 24V / 24Ω = 1 A.

Correct Answer: B — 2 A

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Q. In a circuit with a 24V battery and two resistors of 8 ohms and 16 ohms in series, what is the current flowing through the circuit?

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Solution

Total resistance R = 8 + 16 = 24 ohms. Current I = V/R = 24V / 24Ω = 1 A.

Correct Answer: B — 2 A

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Q. In a circuit with a 24V battery and two resistors of 8 ohms and 4 ohms in series, what is the current flowing through the circuit?

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Solution

Total resistance R_total = 8 + 4 = 12 ohms. Current I = V/R_total = 24V / 12Ω = 2 A.

Correct Answer: B — 2 A

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Q. In a circuit with a 24V battery and two resistors of 8Ω and 16Ω in series, what is the voltage across the 16Ω resistor?

A. 8V
B. 12V
C. 16V
D. 24V

Solution

Voltage across R2 = (R2/(R1+R2)) * V = (16/(8+16)) * 24 = 16V.

Correct Answer: B — 12V

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Q. In a circuit with a 24V battery and two resistors of 8Ω and 4Ω in series, what is the voltage across the 4Ω resistor?

A. 6V
B. 8V
C. 12V
D. 16V

Solution

Using the voltage divider rule, V_R2 = (R2 / (R1 + R2)) * Vtotal = (4 / (8 + 4)) * 24 = 8V.

Correct Answer: C — 12V

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Q. In a circuit with a 24V battery and two resistors of 8Ω and 4Ω in series, what is the current flowing through the circuit?

A. 2A
B. 3A
C. 4A
D. 6A

Solution

Total resistance R_total = 8Ω + 4Ω = 12Ω. Current I = V/R_total = 24V / 12Ω = 2A.

Correct Answer: B — 3A

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Q. In a circuit with a 5V battery and two resistors in parallel (5Ω and 10Ω), what is the current through the 10Ω resistor?

A. 0.5A
B. 1A
C. 1.5A
D. 2A

Solution

Using Ohm's Law, I = V/R = 5V / 10Ω = 0.5A.

Correct Answer: B — 1A

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Q. In a circuit with a 6Ω and a 3Ω resistor in parallel, what is the equivalent resistance?

A. 2Ω
B. 4Ω
C. 1.5Ω
D. 9Ω

Solution

Using the formula for parallel resistors, 1/R_eq = 1/6 + 1/3 = 1/2, R_eq = 2Ω.

Correct Answer: A — 2Ω

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Q. In a circuit with a 9 V battery and a 3 ohm resistor, what is the current flowing through the circuit?

A. 3 A
B. 2 A
C. 1 A
D. 4 A

Solution

Using Ohm's law, I = V / R = 9 V / 3 ohms = 3 A.

Correct Answer: B — 2 A

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Q. In a circuit with a 9V battery and three resistors (3Ω, 3Ω, and 3Ω) in series, what is the voltage across each resistor?

A. 3V
B. 6V
C. 9V
D. 1V

Solution

In a series circuit, the voltage is divided equally among the resistors. Therefore, V across each 3Ω resistor = 9V / 3 = 3V.

Correct Answer: A — 3V

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Q. In a circuit with a 9V battery and three resistors in series (2Ω, 3Ω, and 5Ω), what is the total resistance?

A. 5Ω
B. 8Ω
C. 10Ω
D. 15Ω

Solution

Total resistance R = R1 + R2 + R3 = 2Ω + 3Ω + 5Ω = 10Ω.

Correct Answer: C — 10Ω

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Q. In a circuit with a 9V battery and three resistors of 3Ω, 3Ω, and 6Ω in parallel, what is the total current supplied by the battery?

A. 1A
B. 2A
C. 3A
D. 4A

Solution

1/R_eq = 1/3 + 1/3 + 1/6 = 1. Therefore, R_eq = 1Ω. Total current I = V/R_eq = 9V/1Ω = 9A.

Correct Answer: C — 3A

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Q. In a circuit with a 9V battery and two resistors (R1 = 3Ω, R2 = 6Ω) in series, what is the current flowing through the circuit?

A. 1A
B. 2A
C. 3A
D. 0.5A

Solution

Total resistance R_total = R1 + R2 = 3Ω + 6Ω = 9Ω. Current I = V/R = 9V/9Ω = 1A.

Correct Answer: B — 2A

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Q. In a circuit with a 9V battery and two resistors in series (R1 = 3Ω, R2 = 6Ω), what is the voltage across R2?

A. 6V
B. 3V
C. 9V
D. 0V

Solution

Using voltage division, V_R2 = (R2 / (R1 + R2)) * V_total = (6Ω / (3Ω + 6Ω)) * 9V = 6V.

Correct Answer: A — 6V

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Q. In a circuit with a 9V battery and two resistors of 3 ohms and 6 ohms in parallel, what is the equivalent resistance?

A. 2 ohms
B. 3 ohms
C. 4 ohms
D. 1.5 ohms

Solution

1/R_eq = 1/R1 + 1/R2 = 1/3 + 1/6 = 1/2, R_eq = 2 ohms.

Correct Answer: A — 2 ohms

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Q. In a circuit with a 9V battery and two resistors of 3 ohms and 6 ohms in series, what is the voltage drop across the 6 ohm resistor?

A. 6V
B. 3V
C. 9V
D. 4.5V

Solution

The total resistance is 9 ohms. The current is I = V/R = 9V / 9 ohms = 1 A. The voltage drop across the 6 ohm resistor is V = I * R = 1 A * 6 ohms = 6V.

Correct Answer: A — 6V

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