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

Step 1: Understanding the Concept:
A real battery consists of an ideal electromotive force (EMF) $E$ in series with an internal resistance $r$.
When connected to an external resistance $R$, the current $i$ flowing through the circuit obeys Ohm's Law.
Step 2: Key Formula or Approach:
The total resistance of the circuit is $R + r$.
Applying Ohm's Law:
\[ E = i(R + r) \]
Step 3: Detailed Explanation:
We are given two different conditions to set up a system of linear equations.
Case 1: $R = 5\Omega, i = 1\text{A}$
\[ E = 1 \times (5 + r) \]
\[ E = 5 + r \quad \dots \text{(1)} \]
Case 2: $R = 2\Omega, i = 2\text{A}$
\[ E = 2 \times (2 + r) \]
\[ E = 4 + 2r \quad \dots \text{(2)} \]
Since the EMF $E$ of the cell is constant, equate (1) and (2):
\[ 5 + r = 4 + 2r \]
Rearranging terms to solve for $r$:
\[ 5 - 4 = 2r - r \]
\[ 1 = r \]
Thus, the internal resistance $r$ is 1 $\Omega$.
Step 4: Final Answer:
The internal resistance r is 1 ohm.