The current i in the circuit will be: | vedclass

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Question

(B) $1$. Analyze the circuit: The circuit consists of a $5 \, V$ battery and two parallel branches connected in series with a $20 \, \Omega$ resistor.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) $1$. Analyze the circuit: The circuit consists of a $5 \, V$ battery and two parallel branches connected in series with a $20 \, \Omega$ resistor.$2$. Check diode bias: In the upper branch,the diode is reverse-biased (the $p$-side is connected to the negative terminal relative to the $n$-side),so it acts as an open circuit (no current flows).$3$. In the lower branch,the diode is forward-biased (the $p$-side is connected to the positive terminal),so it acts as a closed switch with negligible resistance.$4$. Calculate total resistance: The total resistance in the active path is the sum of the $20 \, \Omega$ resistor in the main line and the $30 \, \Omega$ resistor in the lower branch: $R_{total} = 20 \, \Omega + 30 \, \Omega = 50 \, \Omega$.$5$. Calculate current: Using Ohm's law,$i = \frac{V}{R_{total}} = \frac{5 \, V}{50 \, \Omega} = \frac{5}{50} \, A$.

The current $i$ in the circuit will be:

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