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$$ I_3 $$ 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 $$ \dfrac{150}{2} = 75 \Omega $$

$$ \therefore \, $$ Net resistance in the circuit = $$ (75 + 25) \Omega = 100 \, \Omega $$

$$ \therefore \, $$ Current $$ I_1 = \dfrac{5}{100} = 0.05 \,A $$

AS resistance of AB and EF are equal and $$ I_1 = I_2 + I_3 + I_4 , I_3 = 0 $$

$$ \therefore \, I_2 = I_4 = \dfrac{0.05}{2} = 0.025\,A $$

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