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In a circuit with a 12V battery and two resistors in series (3 ohms and 6 ohms),

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Question: 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?

Options:

  1. 8 V
  2. 4 V
  3. 6 V
  4. 12 V

Correct Answer: 8 V

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.

In a circuit with a 12V battery and two resistors in series (3 ohms and 6 ohms),

Practice Questions

Q1
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?
  1. 8 V
  2. 4 V
  3. 6 V
  4. 12 V

Questions & Step-by-Step Solutions

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?
Correct Answer: 8 V
  • Step 1: Identify the total resistance in the circuit. You have two resistors in series: 3 ohms and 6 ohms. Add them together: 3 ohms + 6 ohms = 9 ohms.
  • Step 2: Use Ohm's Law to find the current in the circuit. Ohm's Law states that I = V / R. Here, V is the total voltage (12V) and R is the total resistance (9 ohms). So, I = 12V / 9 ohms = 4/3 A.
  • Step 3: Calculate the voltage drop across the 6 ohm resistor using Ohm's Law again. The formula is V = I * R. You already found I (4/3 A) and R for the 6 ohm resistor is 6 ohms. So, V = (4/3 A) * 6 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.
  • Voltage Division – In a series circuit, the voltage drop across a resistor can be calculated using the current flowing through it and its resistance.
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