NCERT Physics I

Question

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Edvaya · Accepted Answer

The correct answer is Subjective. Given: EMF of the battery, $E$ = 8.0 V Internal resistance, $r$ = 0.5 Ω DC supply voltage, $V$ = 120 V Series resistance, $R$ = 15.5 Ω During charging, the effective voltage driving the current is the difference between the supply voltage and the battery EMF. Effective voltage = $V - E = 120 - 8.0 = 112$ V Total resistance of the circuit = $R + r = 15.5 + 0.5 = 16.0$ Ω The charging current $I$ is: $I = \frac{V - E}{R + r} = \frac{112}{16.0} = 7.0$ A The terminal voltage $V_t$ of the battery during charging is: $V_t = E + I r = 8.0 + (7.0 	imes 0.5) = 8.0 + 3.5 = 11.5$ V **Purpose of series resistor:** The series resistor limits the charging current to a safe value. In its absence, the current drawn from the external source would be dangerously high ($I = \frac{112}{0.5} = 224$ A), which could easily damage the battery.

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