Question:
A \(8\ \Omega\) resistor is connected to a battery that has an internal resistance of \(0.2\ \Omega\). If the voltage across the battery, that is the terminal voltage, is \(10\ \text{V}\), then the emf of the battery is
A \(8\ \Omega\) resistor is connected to a battery that has an internal resistance of \(0.2\ \Omega\). If the voltage across the battery, that is the terminal voltage, is \(10\ \text{V}\), then the emf of the battery is
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For a battery delivering current,
\[
E=V+Ir,
\]
where \(V\) is terminal voltage and \(r\) is internal resistance. The emf is greater than terminal voltage during discharge.
Updated On: Jun 26, 2026
- \(10.15\ \text{V}\)
- \(10.20\ \text{V}\)
- \(10.25\ \text{V}\)
- \(9.80\ \text{V}\)
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The Correct Option is C
Solution and Explanation
Step 1: Write the given data.
External resistance is \[ R=8\ \Omega. \] Internal resistance of the battery is \[ r=0.2\ \Omega. \] Terminal voltage is \[ V=10\ \text{V}. \]
Step 2: Find the current through the external resistor.
The terminal voltage appears across the external resistor.
So, \[ V=IR. \] Therefore, \[ I=\frac{V}{R}. \] \[ I=\frac{10}{8}. \] \[ I=1.25\ \text{A}. \]
Step 3: Use the relation between emf and terminal voltage.
For a discharging battery, \[ E=V+Ir, \] where \(E\) is the emf of the battery.
Substituting the values, \[ E=10+(1.25)(0.2). \] \[ E=10+0.25. \] \[ E=10.25\ \text{V}. \]
Step 4: Final conclusion.
Therefore, the emf of the battery is \[ \boxed{10.25\ \text{V}} \] Hence, the correct option is \[ \boxed{(3)} \]
External resistance is \[ R=8\ \Omega. \] Internal resistance of the battery is \[ r=0.2\ \Omega. \] Terminal voltage is \[ V=10\ \text{V}. \]
Step 2: Find the current through the external resistor.
The terminal voltage appears across the external resistor.
So, \[ V=IR. \] Therefore, \[ I=\frac{V}{R}. \] \[ I=\frac{10}{8}. \] \[ I=1.25\ \text{A}. \]
Step 3: Use the relation between emf and terminal voltage.
For a discharging battery, \[ E=V+Ir, \] where \(E\) is the emf of the battery.
Substituting the values, \[ E=10+(1.25)(0.2). \] \[ E=10+0.25. \] \[ E=10.25\ \text{V}. \]
Step 4: Final conclusion.
Therefore, the emf of the battery is \[ \boxed{10.25\ \text{V}} \] Hence, the correct option is \[ \boxed{(3)} \]
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