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AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Physics

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Question : 107 of 160

Marks: +1, -0

The current

i

in the circuit given below is

$-\frac{3E}{4R}$
$-\frac{2E}{R}$
$-\frac{E}{3R}$
$-\;\frac{E}{R}$

Solution:

Now, we apply Kirchhoff's loop rule in loop 1 and loop 2 to get following equations:

In loop 1

0.

-i R - (i+i_1) R - 2E + E = 0

or

-2i R - i_1 R = E

or

i_1 R + 2 i R = -E

\dots(i)

In loop 2

-3E + i_1 R + \frac{i_1}{2} R + i_1 R + 2E + (i+i_1) R = 0

or

\frac{7}{2} i_1 R + i R = E

or

7 i_1 R + 2 i R = 2E

(ii)

Now

7 \times \text{Eq. (i)} - \text{Eq. (ii)}

gives :

7 i_1 R + 14 i R = -7E

7 i_1 R + 2 i R = 2E

\Rightarrow 12 i R = -9E

\Rightarrow i = \frac{-9E}{12R}

\Rightarrow i = \frac{-3E}{4R}

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