Question

Two identical cells each of emf 1.5 V are connected in series across a 10Ω resistance. An ideal voltmeter connected across 10 resistance reads 1.5 V. The internal resistance of each cell is ______Ω.

Answer 1

Correct Answer: 5

Given: 

• EMF of each cell (\( \varepsilon \)) = \( 1.5 \, \text{V} \)
• External resistance (\( R \)) = \( 10 \, \Omega \)
• Voltmeter reading across \( R \): \( V_R = 1.5 \, \text{V} \).

Step 1: Calculate the Current in the Circuit

The cells are connected in series, so:

• Total EMF: \( 2\varepsilon = 2 \times 1.5 = 3 \, \text{V} \)
• Total internal resistance: \( 2r \) (where \( r \) is the internal resistance of each cell).

The current \( I \) in the circuit is given by:

\[ I = \frac{2\varepsilon}{R + 2r}. \]

The voltmeter measures the potential difference across the external resistance \( R \):

\[ V_R = I \cdot R. \]

Rearranging to find \( I \):

\[ I = \frac{V_R}{R}. \]

Substitute \( V_R = 1.5 \, \text{V} \) and \( R = 10 \, \Omega \):

\[ I = \frac{1.5}{10} = 0.15 \, \text{A}. \]

Step 2: Solve for the Internal Resistance

Using the formula for current in the circuit:

\[ I = \frac{2\varepsilon}{R + 2r}. \]

Substitute \( I = 0.15 \, \text{A} \), \( 2\varepsilon = 3 \, \text{V} \), and \( R = 10 \, \Omega \):

\[ 0.15 = \frac{3}{10 + 2r}. \]

Rearranging for \( 10 + 2r \):

\[ 10 + 2r = \frac{3}{0.15} = 20. \]

Simplify for \( 2r \):

\[ 2r = 20 - 10 = 10. \]

Finally, solve for \( r \):

\[ r = \frac{10}{2} = 5 \, \Omega. \]

Final Answer:

The internal resistance of each cell is \( 5 \, \Omega \).