In the following network, the current flowing | vedclass

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(B) Let the nodes be $A, B, C, D$ as shown in the figure. The total current entering the network at node $A$ is $I = 2.1 	ext{ A}$.Let $I_1$ be the c

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$0.9$

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(B) Let the nodes be $A, B, C, D$ as shown in the figure. The total current entering the network at node $A$ is $I = 2.1 	ext{ A}$.Let $I_1$ be the current flowing through the $15 	ext{ } \Omega$ resistor (branch $AB$) and $I_2$ be the current flowing through the $20 	ext{ } \Omega$ resistor (branch $AD$).By Kirchhoff's Current Law at node $A$, $I_1 + I_2 = 2.1 	ext{ A}$.Using Kirchhoff's Voltage Law for the loops:For loop $ABDA$: $15I_1 + 6I_G - 20I_2 = 0$, where $I_G$ is the current through the galvanometer ($6 	ext{ } \Omega$ resistance).For loop $BCDB$: $3(I_1 - I_G) - 4(I_2 + I_G) - 6I_G = 0 \implies 3I_1 - 4I_2 - 13I_G = 0$.Solving these equations for $I_1$ with the condition that the bridge is not balanced, we find the potential at $B$ and $D$ relative to $A$. Alternatively, using nodal analysis: Let $V_A = V$, $V_C = 0$. Let $V_B$ and $V_D$ be the potentials at $B$ and $D$.$(V - V_B)/15 = (V_B - V_C)/3 + (V_B - V_D)/6$$(V - V_D)/20 = (V_D - V_C)/4 + (V_D - V_B)/6$Solving this system for $V_B$ and $V_D$ with $I = (V - V_B)/15 + (V - V_D)/20 = 2.1$, we get $I_1 = 0.9 	ext{ A}$.

In the following network, the current flowing through $15 \text{ } \Omega$ resistance is (in $\text{ A}$)

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