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(D) The given circuit is a Wheatstone bridge. Let the resistances be $P=4 \ \Omega$,$Q=4 \ \Omega$,$R=1 \ \Omega$,and $S=3 \ \Omega$. For a balanced W

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flow from $S$ to $Q$

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(D) The given circuit is a Wheatstone bridge. Let the resistances be $P=4 \ \Omega$,$Q=4 \ \Omega$,$R=1 \ \Omega$,and $S=3 \ \Omega$. For a balanced Wheatstone bridge,the ratio of resistances in opposite arms should be equal,i.e.,$P/R = Q/S$. Here,$4/1 = 4$ and $4/3 = 1.33$. Since $4 
eq 1.33$,the bridge is unbalanced. Let the potential at $P$ be $V_P$ and at $R$ be $V_R$. The potential at $Q$ and $S$ will be determined by the voltage divider rule. $V_Q = V \cdot \frac{4}{4+4} = V/2$ and $V_S = V \cdot \frac{3}{1+3} = 3V/4$. Since $V_S > V_Q$,the current flows from the higher potential to the lower potential,i.e.,from $S$ to $Q$.

In the following network,the current through the galvanometer will:

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