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(D) Using nodal analysis at the junction point above resistor $R$,let the potential be $V_x$. The current $I$ flowing through $R$ is $I = V_x / R$.App

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$25$

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(D) Using nodal analysis at the junction point above resistor $R$,let the potential be $V_x$. The current $I$ flowing through $R$ is $I = V_x / R$.Applying Kirchhoff's Current Law $(KCL)$ at the junction:$(V_x - V_1) / R_1 + (V_x - V_2) / R_2 + V_x / R = 0$$V_x (1/R_1 + 1/R_2 + 1/R) = V_1/R_1 + V_2/R_2$$V_x = \frac{V_1/R_1 + V_2/R_2}{1/R_1 + 1/R_2 + 1/R} = \frac{V_1 R_2 + V_2 R_1}{R_1 R_2 / R_1 + R_1 R_2 / R_2 + R_1 R_2 / R} = \frac{V_1 R_2 + V_2 R_1}{R_2 + R_1 + R_1 R_2 / R}$Since $I = V_x / R$,we have $I = \frac{V_1 R_2 + V_2 R_1}{R(R_1 + R_2) + R_1 R_2}$.Case $1$: $V_1 = 2 \,V, V_2 = 0 \,V, I = 3 \,mA \Rightarrow 3 = \frac{2 R_2}{R(R_1 + R_2) + R_1 R_2} \quad \dots(1)$Case $2$: $V_1 = 0 \,V, V_2 = 4 \,V, I = 4 \,mA \Rightarrow 4 = \frac{4 R_1}{R(R_1 + R_2) + R_1 R_2} \quad \dots(2)$Dividing $(1)$ by $(2)$: $3/4 = (2 R_2) / (4 R_1) \Rightarrow 3/4 = R_2 / (2 R_1) \Rightarrow R_2 / R_1 = 3/2 \Rightarrow R_2 = 1.5 R_1$.Substitute $R_2 = 1.5 R_1$ into $(1)$: $3 = \frac{2(1.5 R_1)}{R(R_1 + 1.5 R_1) + R_1(1.5 R_1)} = \frac{3 R_1}{2.5 R R_1 + 1.5 R_1^2} = \frac{3}{2.5 R + 1.5 R_1}$.So,$2.5 R + 1.5 R_1 = 1$.Case $3$: $V_1 = 10 \,V, V_2 = 10 \,V$. Then $I = \frac{10 R_2 + 10 R_1}{R(R_1 + R_2) + R_1 R_2} = \frac{10(R_1 + R_2)}{R(R_1 + R_2) + R_1 R_2}$.Using $R_2 = 1.5 R_1$,$I = \frac{10(2.5 R_1)}{R(2.5 R_1) + 1.5 R_1^2} = \frac{25 R_1}{2.5 R R_1 + 1.5 R_1^2} = \frac{25}{2.5 R + 1.5 R_1}$.Since $2.5 R + 1.5 R_1 = 1$,$I = 25 / 1 = 25 \,mA$.

Two batteries $V_1$ and $V_2$ are connected to three resistors as shown below. If $V_1=2 \,V$ and $V_2=0 \,V$,then the current $I=3 \,mA$. If $V_1=0 \,V$ and $V_2=4 \,V$,then the current $I=4 \,mA$. Now,if $V_1=10 \,V$ and $V_2=10 \,V$,then the current $I$ will be ............ $\,mA$.

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