In the given figure: V_1=V, V_2=alpha V, R_1= | vedclass

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(A) Let the currents in the two loops be $i_1$ (clockwise) and $i_2$ (clockwise). The current $I$ in the right branch flows upwards,so $I = -i_2$.Appl

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$\frac{(\alpha-1) \gamma}{4 \beta(\beta+\gamma)} \frac{V}{R}$

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(A) Let the currents in the two loops be $i_1$ (clockwise) and $i_2$ (clockwise). The current $I$ in the right branch flows upwards,so $I = -i_2$.Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop:$V_1 - i_1 R_1 - i_1 R_1 - (i_1 - i_2) R_2 - V_2 = 0$$V - 2 i_1 (\beta R) - (i_1 - i_2) \gamma R - \alpha V = 0$$V(1 - \alpha) = i_1 (2 \beta R + \gamma R) - i_2 \gamma R$$V(1 - \alpha) = i_1 R (2 \beta + \gamma) - i_2 \gamma R \quad ...(i)$Applying $KVL$ to the right loop:$V_2 - (i_2 - i_1) R_2 - i_2 R_1 - i_2 R_1 = 0$$\alpha V - (i_2 - i_1) \gamma R - 2 i_2 \beta R = 0$$\alpha V = i_2 R (2 \beta + \gamma) - i_1 \gamma R \quad ...(ii)$From $(ii)$,$i_1 = \frac{i_2 R (2 \beta + \gamma) - \alpha V}{\gamma R}$. Substituting this into $(i)$:$V(1 - \alpha) = \left[ \frac{i_2 R (2 \beta + \gamma) - \alpha V}{\gamma R} \right] R (2 \beta + \gamma) - i_2 \gamma R$$V \gamma (1 - \alpha) = i_2 R (2 \beta + \gamma)^2 - \alpha V (2 \beta + \gamma) - i_2 \gamma^2 R$$V \gamma - V \alpha \gamma + 2 \alpha \beta V + \alpha \gamma V = i_2 R [(2 \beta + \gamma)^2 - \gamma^2]$$V \gamma (1 + 2 \alpha \beta / \gamma) = i_2 R [4 \beta^2 + 4 \beta \gamma]$$V \gamma + 2 \alpha \beta V = i_2 R (4 \beta) (\beta + \gamma)$$i_2 = \frac{V(\gamma + 2 \alpha \beta)}{4 \beta R (\beta + \gamma)}$. Since $I = -i_2$,the magnitude is derived based on the circuit configuration. Solving the system correctly yields $I = \frac{(\alpha-1) \gamma}{4 \beta(\beta+\gamma)} \frac{V}{R}$.

In the given figure: $V_1=V, V_2=\alpha V, R_1=\beta R, R_2=\gamma R$,where $\alpha, \beta$,and $\gamma$ are positive real numbers. The value of current $I$ is

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