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In a circuit with a 9V battery and two resistors of 3 ohms and 6 ohms in series,

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Question: 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?

Options:

  1. 6V
  2. 3V
  3. 9V
  4. 4.5V

Correct Answer: 6V

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.

In a circuit with a 9V battery and two resistors of 3 ohms and 6 ohms in series,

Practice Questions

Q1
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?
  1. 6V
  2. 3V
  3. 9V
  4. 4.5V

Questions & Step-by-Step Solutions

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?
  • Step 1: Identify the total resistance in the circuit. Add the resistances of the two resistors: 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 voltage of the battery (9V) and R is the total resistance (9 ohms). So, I = 9V / 9 ohms = 1 A.
  • Step 3: Calculate the voltage drop across the 6 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 6 ohm resistor. So, V = 1 A * 6 ohms = 6V.
  • 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.
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