Question 52: Current Electricity

Question

A wire of uniform cross section and resistance 4 ohms is bent in the shape of square ABCD. Point A is connected to a point $$\mathrm{P}$$ on DC by a wire AP of resistance $$1 \mathrm{ohm}$$. When a potential difference is applied between $$\mathrm{A}$$ and $$\mathrm{C}$$, the points $$\mathrm{B}$$ and $$\mathrm{P}$$ are seen to be in same potential. What is the resistance of part $$\mathrm{DP}$$ ?

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

$$
\sqrt{2}-1
$$ Ohm

Answer

2 Ohm

Answer

1 Ohm

Answer

Zero Ohm

Current Electricity

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