A network of resistors is connected as shown in the figure. The potential drop across 4Ω resistors is

2 V

The correct answer is Option (1) → 2 V

Equivalent resistance of left pair: $R_L=\frac{4\times4}{4+4}=2\ \Omega$

Equivalent resistance of right pair: $R_R=\frac{12\times6}{12+6}=4\ \Omega$

Total resistance: $R_{\text{tot}}=R_L+1+R_R+1=2+1+4+1=8\ \Omega$

Current: $I=\frac{8}{8}=1\ \text{A}$

Potential drop across each 4 Ω (they share the left equivalent 2 Ω): $V=I\times R_L=1\times2=2\ \text{V}$

Answer: $2\ \text{V}$