Question 53: Current Electricity

Question

A storage battery of emf $$28.0 \mathrm{~V}$$ and internal resistance $$0.5 \Omega$$ is being charged by a $$140 \mathrm{~V}$$ dc supply using a series resistor of $$27.5 \Omega$$. The terminal voltage of the battery during charging is

Accepted Answer

Let's analyze the problem step by step.
We have a charging circuit that includes:
A supply voltage: $$140\,	ext{V}$$
A series resistor: $$R = 27.5\,\Omega$$
A battery with:
Electromotive force (emf): $$\mathcal{E} = 28.0\,	ext{V}$$
Internal resistance: $$r = 0.5\,\Omega$$
Write the loop equation using KirchhoffÃ¢ÂÂs Voltage Law (KVL). Since the battery is being charged, its emf opposes the charging current, so the KVL equation is:
$$140 - I\,(27.5) - I\,(0.5) - 28 = 0$$
Combine the resistor drops:
$$140 - I\,(27.5 + 0.5) - 28 = 0$$
$$140 - I\,(28) - 28 = 0$$
Isolate the current $$I$$:
$$140 - 28 = I 	imes 28$$
$$112 = 28I$$
$$I = \frac{112}{28} = 4\,	ext{A}$$
The terminal voltage $$V_{	ext{term}}$$ of the battery when charging is the sum of its emf and the voltage drop across its internal resistance:
$$V_{	ext{term}} = \mathcal{E} + I\,r$$
$$V_{	ext{term}} = 28.0\,	ext{V} + 4\,	ext{A} 	imes 0.5\,\Omega$$
$$V_{	ext{term}} = 28.0\,	ext{V} + 2.0\,	ext{V} = 30.0\,	ext{V}$$
Thus, the terminal voltage of the battery during charging is $$\mathbf{30\,V}$$, which corresponds to Option D.

Answer

26V

Answer

2V

Answer

5V

Answer

30V

Current Electricity

A storage battery of emf $28.0 \mathrm{~V} and internal resistance 0.5 \Omega is being charged by a 140 \mathrm{~V} dc supply using a series resistor of 27.5 \Omega$. The terminal voltage of the battery during charging is

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