In the circuit shown E - F - G and H are cells of e-m-f- 2 V - 1 V - 3 V and 1 V respectively and their internal resistances are 2 - 1 - 3 - and 1 - respectively-A- V D - V B -2-13 VB- V D - V B -2-13 VC- V G -21-13 V - potential difference across GD- V H -19-13 V - potential difference across H

The correct options are A V_D V_B= 213 V C V_G=2113 V= potential difference across G D V_H = 1913 V= potential difference across H Let potential of point B be x ...

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Byju's Answer

Standard XII

Physics

RLC Circuit

In the circui...

Question

In the circuit shown $E, F, G$ and $H$ are cells of e.m.f. $2 V, 1 V, 3 V$ and $1 V$ respectively and their internal resistances are $2 Ω, 1 Ω, 3 Ω$ and $1 Ω$ respectively.

V_{D} - V_{B} = - \frac{2}{13} V

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V_{D} - V_{B} = \frac{2}{13} V

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V_{G} = \frac{21}{13} V =

potential difference across

G

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V_{H} = \frac{19}{13} V =

potential difference across

H

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Solution

The correct options are
A $V_{D} - V_{B} = - \frac{2}{13} V$
C $V_{G} = \frac{21}{13} V =$ potential difference across $G$
D $V_{H} = \frac{19}{13} V =$ potential difference across $H$

Let potential of point $B$ be $x$ and at point $D$ be $0$ , then from KCL at $D$
$i_{1} + i_{2} + i_{3} = 0$
$\frac{x}{2} + \frac{x - 2}{4} + \frac{x + 1}{3} = 0$
$= \frac{6 x + 3 x - 6 + 4 x + 4}{12} = 0$
$\Rightarrow 13 x = 2$
$x = \frac{2}{13} volt$
Putting the value of $x$ , we get $i_{2} = \frac{- 6}{13}$
Now using KVL and putting $i_{2} = \frac{- 6}{13}$ ,
$V_{G} = 3 - \frac{6}{13} (3) = \frac{21}{13} V$ and
$V_{H} = 1 + \frac{6}{13} (1) = \frac{19}{13} V$
$V_{D} - V_{B} = - \frac{1}{13} (2) = - \frac{2}{13} V$

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In the circuit shown E, F, G and H are cells of e.m.f. 2V, 1V, 3V and 1V respectively and their internal resistances are 2Ω, 1Ω, 3Ω and 1Ω respectively.

In the circuit shown $E, F, G$ and $H$ are cells of e.m.f. $2 V, 1 V, 3 V$ and $1 V$ respectively and their internal resistances are $2 Ω, 1 Ω, 3 Ω$ and $1 Ω$ respectively.