Question

If a cell of 12 V emf delivers 2 A current in a circuit having a resistance of 5.8 \(\Omega\), then the internal resistance of the cell is:

• A. 1 \(\Omega\)
• B. 0.2 \(\Omega\)
• C. 0.3 \(\Omega\)
• D. 0.6 \(\Omega\)
• E. 0.8 \(\Omega\)

Hint:

The internal resistance of a cell can be found using the equation \( E = I \times (R + r) \). If the current and external resistance are known, the internal resistance can be calculated.

Answer 1

The Correct Option is B

We are given the following information: emf of the cell, \(E = 12 \, {V}\) current delivered, \(I = 2 \, {A}\) external resistance, \(R = 5.8 \, \Omega\) 
The total resistance in the circuit \(R_{{total}}\) is the sum of the internal resistance \(r\) and the external resistance \(R\): \[ R_{{total}} = R + r \] Using Ohm's law for the total circuit, we have: \[ E = I \times R_{{total}} = I \times (R + r) \] Substituting the given values: \[ 12 = 2 \times (5.8 + r) \] Solving for \(r\): \[ 12 = 2 \times 5.8 + 2r \quad \Rightarrow \quad 12 = 11.6 + 2r \quad \Rightarrow \quad 2r = 12 - 11.6 = 0.4 \] \[ r = \frac{0.4}{2} = 0.2 \, \Omega \] Thus, the internal resistance of the cell is \(0.2 \, \Omega\).
Therefore, the correct answer is option (B), 0.2 \(\Omega\).