Question:
A storage battery of emf \(10V\) and internal resistance \(1 \Omega\) is being charged by a \(100V\) dc supply using a series resistor of \(17 \Omega\).
- The terminal voltage of the battery during charging is
A storage battery of emf \(10V\) and internal resistance \(1 \Omega\) is being charged by a \(100V\) dc supply using a series resistor of \(17 \Omega\).
- The terminal voltage of the battery during charging is
Show Hint
When a battery is charging, the terminal voltage is given by:
\[
V_T = E + I r
\]
where \(I = \frac{V - E}{R + r}\). This is different from discharging mode, where \(V_T = E - I r\).
Updated On: May 5, 2026
- \(25V\)
- \(30V\)
- \(20V\)
- \(15V\)
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The Correct Option is D
Solution and Explanation
- In a charging circuit, the external source drives current into the battery, so the net driving emf is: \[ V - E \]
- Total resistance in the circuit:
- External resistance = \(17 \Omega\)
- Internal resistance = \(1 \Omega\)
- So total resistance: \[ R_{total} = 17 + 1 = 18 \Omega \]
- Current in the circuit: \[ I = \frac{V - E}{R_{total}} = \frac{100 - 10}{18} \] \[ I = \frac{90}{18} = 5A \]
- For a charging battery, terminal voltage is: \[ V_T = E + Ir \]
- Substitute values: \[ V_T = 10 + (5 \times 1) \] \[ V_T = 15V \]
- Hence, the terminal voltage is: 15 V
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