Question

Figure shows a circuit consisting of cell of emf \(E\) and internal resistance \(r\) connected in series with external resistance \(R\). When \(R = 5\Omega\), current \(i = 1\,A\) and when \(R = 2\Omega\), current \(i = 2\,A\). Find \(r\) (in ohm). 

• A. \(1\Omega\)
• B. \(2\Omega\)
• C. \(3\Omega\)
• D. \(4\Omega\)

Hint:

For circuits with internal resistance: \[ E = I(R+r) \] Use two different current conditions to form simultaneous equations.

Answer 1

The Correct Option is A

Concept: Using Ohm's law for a circuit with internal resistance: \[ E = I(R + r) \]
Step 1: Apply the condition for \(R = 5\Omega\). \[ E = 1(5 + r) \] \[ E = 5 + r \qquad ...(1) \]
Step 2: Apply the condition for \(R = 2\Omega\). \[ E = 2(2 + r) \] \[ E = 4 + 2r \qquad ...(2) \]
Step 3: Solve the equations. \[ 5 + r = 4 + 2r \] \[ r = 1\Omega \] \[ \boxed{r = 1\Omega} \]