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(D) Let us analyze the circuit using Kirchhoff's Voltage Law $(KVL)$.Consider the first loop on the left. It contains two $2 \ V$ batteries connected

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(D) Let us analyze the circuit using Kirchhoff's Voltage Law $(KVL)$.Consider the first loop on the left. It contains two $2 \ V$ batteries connected in opposition to each other and a $1 \ \Omega$ resistor.The net electromotive force $(EMF)$ in this loop is $2 \ V - 2 \ V = 0 \ V$.Since the net $EMF$ is zero,the current flowing through the $1 \ \Omega$ resistor in this loop is $I = V/R = 0/1 = 0 \ A$.Similarly,for the other loops,the batteries are arranged such that their potentials cancel each other out.Therefore,no current flows through any of the resistors in the circuit.The current in each resistance is $0 \ A$.

In the given circuit,the current in each resistance is: (in $A$)

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