The potential difference across 8,Omega resis | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. The circuit consists of a series combination of three main blocks: the $3\,\Omega$ resistor,the upper parallel network,the lower parallel net

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(A) $1$. The circuit consists of a series combination of three main blocks: the $3\,\Omega$ resistor,the upper parallel network,the lower parallel network,and the $1\,\Omega$ resistor.$2$. The lower parallel network contains $24\,\Omega$ and $8\,\Omega$ resistors in parallel. The equivalent resistance $R_{lower}$ is $\frac{24 	imes 8}{24 + 8} = \frac{192}{32} = 6\,\Omega$.$3$. Given the potential difference across the $8\,\Omega$ resistor is $48\,V$,the current $i$ flowing through this parallel block is $i = \frac{48\,V}{6\,\Omega} = 8\,A$.$4$. The upper parallel network consists of $20\,\Omega$,$30\,\Omega$,and $60\,\Omega$ resistors in parallel. The equivalent resistance $R_{upper}$ is $\frac{1}{R_{upper}} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3+2+1}{60} = \frac{6}{60} = \frac{1}{10}$,so $R_{upper} = 10\,\Omega$.$5$. The total resistance between $X$ and $Y$ is $R_{XY} = 3\,\Omega + R_{upper} + R_{lower} + 1\,\Omega = 3 + 10 + 6 + 1 = 20\,\Omega$.$6$. Since the same current $i = 8\,A$ flows through the entire series path,the potential difference between $X$ and $Y$ is $V_{XY} = i 	imes R_{XY} = 8\,A 	imes 20\,\Omega = 160\,V$.

The potential difference across $8\,\Omega$ resistance is $48\,V$ as shown in the figure. The value of potential difference across $X$ and $Y$ points will be ............. $V$.

Similar Questions

Pick out the wrong statement.

Two bulbs consume the same power when operated at $200\, V$ and $300\, V$ respectively. When these bulbs are connected in series across a $D.C.$ source of $500\, V$,then:

Antimony and bismuth are usually used in a thermocouple,because

The expression for thermo $e.m.f.$ in a thermocouple is given by the relation $E = 40\theta - \frac{\theta^2}{20}$,where $\theta$ is the temperature difference of two junctions. For this,the neutral temperature will be ................... $^oC$.

When current is passed in an antimony-bismuth couple,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module