Find the current i in the circuit shown. | vedclass

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(B) In the given circuit,the diode in the upper branch is reverse-biased because its p-side is connected to the negative terminal of the battery (rela

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$\frac{5}{50} \, A$

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(B) In the given circuit,the diode in the upper branch is reverse-biased because its p-side is connected to the negative terminal of the battery (relative to the n-side). Therefore,no current flows through the upper branch.The diode in the middle branch is forward-biased because its p-side is connected to the positive terminal of the battery. Thus,current flows through this branch.The total resistance of the circuit is the sum of the resistor in the middle branch $(30 \, \Omega)$ and the resistor in the main branch $(20 \, \Omega)$.Total resistance $R_{eq} = 30 \, \Omega + 20 \, \Omega = 50 \, \Omega$.Using Ohm's law,the current $i$ is given by:$i = \frac{V}{R_{eq}} = \frac{5 \, V}{50 \, \Omega} = \frac{5}{50} \, A$.

Find the current $i$ in the circuit shown.

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