Question:
The current \(i\) in the network is :
The current \(i\) in the network is :
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Decide whether each ideal diode is forward or reverse biased from the battery polarity. A reverse-biased diode removes its branch from the circuit.
Updated On: Nov 5, 2026
- 0.6 A
- 0.3 A
- 0 A
- 0.2 A
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The Correct Option is B
Solution and Explanation
The 9 V battery has its longer (positive) plate on the left, so the current \(i\) enters the bridge at the left junction and leaves at the right junction.
In the bridge, the left diode (in series with the upper-left 5 \(\Omega\)) and the right diode (in series with the lower-right 20 \(\Omega\)) are both oriented against the current that would flow from the left junction to the right junction. Both are reverse biased and carry no current, so those two branches are open.
The only conducting path is: left junction - 10 \(\Omega\) (lower left) - middle 5 \(\Omega\) - 10 \(\Omega\) (upper right) - right junction.
Bridge resistance = \(10 + 5 + 10 = 25\ \Omega\). Total with the series 5 \(\Omega\) = \(30\ \Omega\).
\(i = \frac{9}{30} = 0.3\) A.
Answer: (2) 0.3 A
In the bridge, the left diode (in series with the upper-left 5 \(\Omega\)) and the right diode (in series with the lower-right 20 \(\Omega\)) are both oriented against the current that would flow from the left junction to the right junction. Both are reverse biased and carry no current, so those two branches are open.
The only conducting path is: left junction - 10 \(\Omega\) (lower left) - middle 5 \(\Omega\) - 10 \(\Omega\) (upper right) - right junction.
Bridge resistance = \(10 + 5 + 10 = 25\ \Omega\). Total with the series 5 \(\Omega\) = \(30\ \Omega\).
\(i = \frac{9}{30} = 0.3\) A.
Answer: (2) 0.3 A
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