In a circuit with a 12 V battery and two resistors in series (4 ohms and 8 ohms)

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What’s inside this PDF?

Question: 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?

Options:

4 V
8 V
6 V
12 V

Correct Answer: 8 V

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.

In a circuit with a 12 V battery and two resistors in series (4 ohms and 8 ohms)

Practice Questions

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?

4 V
8 V
6 V
12 V

Questions & Step-by-Step Solutions

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?

Correct Answer: 8 V

Step 1: Identify the total resistance in the circuit. Add the two resistors together: 4 ohms + 8 ohms = 12 ohms.
Step 2: Use Ohm's Law to find the current in the circuit. The formula is I = V / R. Here, V is the battery voltage (12 V) and R is the total resistance (12 ohms). So, I = 12 V / 12 ohms = 1 A.
Step 3: Calculate the voltage drop across the 8 ohm resistor using Ohm's Law again. The formula is V = I * R. Here, I is the current (1 A) and R is the resistance of the 8 ohm resistor. So, V = 1 A * 8 ohms = 8 V.

Ohm's Law – The relationship between voltage (V), current (I), and resistance (R) in an electrical circuit, expressed as V = I * R.
Series Circuits – In a series circuit, the total resistance is the sum of individual resistances, and the same current flows through all components.

In a circuit with a 12 V battery and two resistors in series (4 ohms and 8 ohms)

What’s inside this PDF?

In a circuit with a 12 V battery and two resistors in series (4 ohms and 8 ohms)

Practice Questions

Questions & Step-by-Step Solutions

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