Question

A battery of emf \(12\,\text{V}\) and internal resistance \(4\,\Omega\) is connected to a resistor. The resistance of the resistor if the current in the circuit is \(0.8\,\text{A}\) is

• A. \(11\,\Omega\)
• B. \(9\,\Omega\)
• C. \(15\,\Omega\)
• D. \(13\,\Omega\)

Hint:

When a cell has internal resistance \(r\), the total resistance in the circuit is \[ R+r. \] Hence, \[ I=\frac{E}{R+r}. \] Always include the internal resistance while applying Ohm's law to a complete circuit.

Answer 1

The Correct Option is A

Step 1: Write Ohm's law with internal resistance.
I = E/(R+r). Rearranging: R = E/I – r.

Step 2: Substitute given values.
R = 12/0.8 – 4 = 15 – 4 = 11 Ω.

Step 3: Verify by back-substitution.
I = 12/(11+4) = 12/15 = 0.8 A.

Step 4: Final Answer:
11 Ω.