Q147 / 160 · AP EAMCET 2022 (07 Jul Shift 1) · Physics
The current $i$ in the circuit given below is
- Correct answerA$-\frac{3 E}{4 R}$
- B$-\frac{2 E}{R}$
- C$-\frac{E}{3 R}$
- D$-\frac{E}{R}$
Solution
Lets assume current distribution in given circuit is as shown
Now, we apply Kirchhoff's loop rule in loop 1 and loop 2 to get following equations: In loop 1, $-i R-\left(i+i_1\right) R-2 E+E=0$ or $\quad-2 i R-i_1 R=E$ or $\quad i_1 R+2 i R=-E$ ...(i) And in loop 2, $-3 E+i_1 R+\frac{i_1}{2} R+i_1 R+2 E+\left(i+i_1\right) R=0$ or $\cdot \frac{7}{2} i_1 R+i R=E$ or $\quad 7 i_1 R+2 i R=2 E$ ...(ii) Now $7 \times$ eq. (i)-eq. (ii) gives
$\Rightarrow \quad 12 i R=-9 E$ $\Rightarrow \quad i=\frac{-9 E}{12 R}$ $\Rightarrow \quad i=\frac{-3 E}{4 R}$
Now, we apply Kirchhoff's loop rule in loop 1 and loop 2 to get following equations: In loop 1, $-i R-\left(i+i_1\right) R-2 E+E=0$ or $\quad-2 i R-i_1 R=E$ or $\quad i_1 R+2 i R=-E$ ...(i) And in loop 2, $-3 E+i_1 R+\frac{i_1}{2} R+i_1 R+2 E+\left(i+i_1\right) R=0$ or $\cdot \frac{7}{2} i_1 R+i R=E$ or $\quad 7 i_1 R+2 i R=2 E$ ...(ii) Now $7 \times$ eq. (i)-eq. (ii) gives
$\Rightarrow \quad 12 i R=-9 E$ $\Rightarrow \quad i=\frac{-9 E}{12 R}$ $\Rightarrow \quad i=\frac{-3 E}{4 R}$
Hence, the correct answer is option: 'A'.