Question

Current in a circuit is 0.6 A when an external resistance of \( 3 \, \Omega \) is connected. When the external resistance is changed to \( 6 \, \Omega \), current in the circuit becomes 0.4 A. Find the internal resistance of the cell.

Hint:

When two equations involving current and resistance are given, use Ohm's law to set up a system of equations and solve for the unknowns.

Answer 1

Step 1: Use the formula for current.
The total resistance in the circuit is the sum of the internal resistance \( r \) and the external resistance \( R \). Using Ohm's law: \[ I = \frac{E}{r + R} \] where \( I \) is the current, \( E \) is the emf of the cell, \( r \) is the internal resistance, and \( R \) is the external resistance.
Step 2: Set up equations for both conditions.
For the first condition, when \( R = 3 \, \Omega \) and \( I = 0.6 \, \text{A} \): \[ 0.6 = \frac{E}{r + 3} \] For the second condition, when \( R = 6 \, \Omega \) and \( I = 0.4 \, \text{A} \): \[ 0.4 = \frac{E}{r + 6} \]
Step 3: Solve the system of equations.
From the first equation: \[ E = 0.6 \times (r + 3) = 0.6r + 1.8 \] From the second equation: \[ E = 0.4 \times (r + 6) = 0.4r + 2.4 \] Equating the two expressions for \( E \): \[ 0.6r + 1.8 = 0.4r + 2.4 \] Solving for \( r \): \[ 0.6r - 0.4r = 2.4 - 1.8 \quad \Rightarrow \quad 0.2r = 0.6 \quad \Rightarrow \quad r = \frac{0.6}{0.2} = 3 \, \Omega \] Thus, the internal resistance of the cell is: \[ \boxed{3 \, \Omega} \]