Q.
In the given circuit, current through $3 \, \Omega $ resistor is
Solution:
Net resistance of the circuit,
$R = 2 + \frac{6 \times 3}{6 +3} = 4 \Omega $
$I = \frac{V}{R} = \frac{6}{4} = 1.5 \, A $
Current through $3 \Omega = \frac{(6 \Omega)(1.5 A)}{(3 \Omega + 6 \Omega)} = 1 A$
