\\\\( I_3 \\\\) is zero as the diode in that branch is reverse biased. Resistance in the branch AB and EF are each \\\\( (125 + 25)\\Omega = 150 \\Omega \\\\) As AB and EF are identical parallel branches , their effective resistance is \\\\( \\dfrac{150}{2} = 75 \\Omega \\\\) \\\\( \ herefore \\, \\\\) Net resistance in the circuit = \\\\( (75 + 25) \\Omega = 100 \\, \\Omega \\\\) \\\\( \ herefore \\, \\\\) Current \\\\( I_1 = \\dfrac{5}{100} = 0.05 \\,A \\\\) AS resistance of AB and EF are equal and \\\\( I_1 = I_2 + I_3 + I_4 , I_3 = 0 \\\\) \\\\( \ herefore \\, I_2 = I_4 = \\dfrac{0.05}{2} = 0.025\\,A \\\\)