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Calculate the net resistance of the circuit between A and B :
A
$\dfrac{8}{3}\ \Omega$
B
$\dfrac{14}{3}\ \Omega$
C
$\dfrac{16}{3}\ \Omega$
D
$\dfrac{22}{3}\ \Omega$
Detailed Solution
The arrangement is a Wheatstone bridge with arms 3, 4 (top) and 6, 8 (bottom), and 7 $\Omega$ in the middle branch.
Check balance: $\dfrac{3}{4} = \dfrac{6}{8}$, so the bridge is balanced and no current flows through the 7 $\Omega$ resistor. It can be removed.
Top branch: $3 + 4 = 7\ \Omega$; bottom branch: $6 + 8 = 14\ \Omega$. These are in parallel.
$R = \dfrac{R_1R_2}{R_1 + R_2} = \dfrac{14 \times 7}{21} = \dfrac{14}{3}\ \Omega$
Check balance: $\dfrac{3}{4} = \dfrac{6}{8}$, so the bridge is balanced and no current flows through the 7 $\Omega$ resistor. It can be removed.
Top branch: $3 + 4 = 7\ \Omega$; bottom branch: $6 + 8 = 14\ \Omega$. These are in parallel.
$R = \dfrac{R_1R_2}{R_1 + R_2} = \dfrac{14 \times 7}{21} = \dfrac{14}{3}\ \Omega$