For the circuit shown in figure, choose then correct options:

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1,The current in $d b$ is $\frac{1}{13}$ amp

2.The $P D$ across $d b$ is $\frac{2}{13}$ volt

3.The $P D$ across the ceil $(C)$ is $\frac{21}{13}$ volt

4.The $P D$ across the cell $(B)$ is $\frac{19}{13}$ volt

answer is B, C, D.

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Detailed Solution

(CD) From the loop, $b a d b 2 i_{1} - 2 + i_{1} + 1 + 2 (i_{1} - i_{2}) = 0$

$5 i_{1} - 2 i_{2} = 1$ ………….. $(1)$

From the loop $d c b d,$

3 - 3 $i_{2}$ - 1 $\times i_{2}$ -1+2( $i_{1} - i_{2}) = 0$

or, 3 $i_{2} - i_{1} = 1 . . . . . . . . . . . (2)$

From eqs. (1)and( 2).

$i_{1} = \frac{5}{13} a m p, a n d i_{2} = \frac{6}{13} a m p$

So, $i_{1} - i_{2} = - \frac{1}{13} a m p$

$V_{b} - V_{d} = \frac{2}{13} v o l t$

The $P D$ across the ceil $(C)$ is

$V_{c} - V_{d} = - 3 i_{2} + 3 = \frac{21}{13} v o l t$

The $P D$ across the ceil $(B)$ is

$V_{c} - V_{b} = i_{2} + 1 = \frac{19}{13} v o l t$

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