Question

The potential difference across $8$ ohm resistance is $48$ volt as shown in the figure. The value of potential difference across X and Y points will be

(A) $128$ volt
(B) $160$ volt
(C) $80$ volt
(D) $62$ volt

Answer 1

Hint : To solve this question, we need to assume a battery between the points X and Y of emf equal to the unknown potential difference. Then we have to simplify the circuit by calculating the parallel equivalent resistance.

Complete step by step answer
Let the potential difference across X and Y points be $V$ . So let us consider a battery of emf $V$ connected across the points X and Y as shown in the figure below.

According to the question, the potential difference across the $8$ ohm resistance is equal to $48$ volt. So referring to the above figure, the potential difference between the points A and B is $48$ volt. According to the polarity of the battery, the point B should be at a higher potential than the point A. So we have
$\Rightarrow {V_{BA}} = 8{\text{V}}$ ……………………….(1)
Now, the $24\Omega$ and $8\Omega$ resistances are in parallel combination across the point A and B. So the equivalent resistance between the points A and B, ${R_{AB}}$ is given by
$\Rightarrow \dfrac{1}{{{R_{AB}}}} = \dfrac{1}{8} + \dfrac{1}{{24}}$
On solving we get
$\Rightarrow {R_{AB}} = 6\Omega$
Also, across the points B and C, the $20\Omega ,30\Omega ,60\Omega$ resistances are arranged in parallel combination. So the equivalent resistance across the points B and C is given by
$\Rightarrow \dfrac{1}{{{R_{BC}}}} = \dfrac{1}{{20}} + \dfrac{1}{{30}} + \dfrac{1}{{60}}$
On solving we get
$\Rightarrow {R_{BC}} = 10\Omega$
So the above circuit can be redrawn as

Now, let $I$ be the current in the circuit. From Ohm’s law we have
$\Rightarrow V = IR$ ..........................(2)
Now, the equivalent resistance in the circuit is given by
$\Rightarrow R = 3\Omega + 10\Omega + 6\Omega + 1\Omega$
$\Rightarrow R = 20\Omega$
So from (2) we have
$\Rightarrow V = 20I$
$\Rightarrow I = \dfrac{V}{{20}}$ ........................(3)
The potential difference across the points A and B is equal to that across the $6\Omega$ resistance. Therefore we have
$\Rightarrow {V_{BA}} = 6I$
From (3)
$\Rightarrow {V_{BA}} = \dfrac{{6V}}{{20}}$ ........................(4)
Equating (1) and (4) we have
$\Rightarrow \dfrac{{6V}}{{20}} = 48$
$\Rightarrow V = 160{\text{V}}$
Thus, the potential difference between the points X and Y is equal to $160$ volts.
Hence, the correct answer is option B.

Note
The points X and Y are not shown to be connected through any battery in the figure given in the question. But the potential difference which is given across the $8$ ohm resistance clearly indicates that a voltage source must be present there. Hence we were able to assume the battery between X and Y.