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 the relation for current in a circuit with internal resistance.
For a cell of emf \(E\), internal resistance \(r\), and external resistance \(R\), \[ I=\frac{E}{R+r} \] Given, \[ E=12\,\text{V} \] \[ r=4\,\Omega \] \[ I=0.8\,\text{A} \]

Step 2: Substitute the given values.
Using \[ I=\frac{E}{R+r}, \] we get \[ 0.8=\frac{12}{R+4} \] Cross-multiplying, \[ 0.8(R+4)=12 \] \[ R+4=\frac{12}{0.8} \] \[ R+4=15 \]

Step 3: Calculate the resistance.
\[ R=15-4 \] \[ R=11\,\Omega \]

Step 4: Verification.
Substituting \(R=11\,\Omega\), \[ I=\frac{12}{11+4} \] \[ I=\frac{12}{15} \] \[ I=0.8\,\text{A} \] which matches the given current.

Step 5: Final conclusion.
Therefore, the resistance of the resistor is \[ \boxed{11\,\Omega} \]