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(C) Let the potential at point $D$ be $V_D$. Applying Kirchhoff's Current Law $(KCL)$ at node $D$,the sum of currents leaving the node must be zero:$I

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currents in the paths $AD$,$DB$ and $DC$ are in the ratio of $3: 2: 1$

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(C) Let the potential at point $D$ be $V_D$. Applying Kirchhoff's Current Law $(KCL)$ at node $D$,the sum of currents leaving the node must be zero:$I_{AD} + I_{DB} + I_{DC} = 0$$\frac{V_D - 70}{10} + \frac{V_D - 0}{20} + \frac{V_D - 10}{30} = 0$Multiplying by $60$ to clear the denominators:$6(V_D - 70) + 3(V_D) + 2(V_D - 10) = 0$$6V_D - 420 + 3V_D + 2V_D - 20 = 0$$11V_D = 440$$V_D = 40 \,V$Now,calculate the currents:$I_{AD} = \frac{70 - 40}{10} = 3 \,A$ (current flows from $A$ to $D$)$I_{DB} = \frac{40 - 0}{20} = 2 \,A$ (current flows from $D$ to $B$)$I_{DC} = \frac{40 - 10}{30} = 1 \,A$ (current flows from $D$ to $C$)Thus,the ratio of currents in paths $AD$,$DB$,and $DC$ is $3: 2: 1$.

In the circuit given here,the potentials at points $A$,$B$,and $C$ are $70 \,V$,$0 \,V$,and $10 \,V$ respectively. Then:

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