The circuit shown in the figure contains two diodes- each with a forward resistance of 50 - and with infinite backward resistance- If the battery voltage is 6 V- the current through the 100- - resistance is

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Standard XII

Physics

V-I Characteristics for a P-N Junction Diode

The circuit s...

Question

The circuit shown in the figure contains two diodes, each with a forward resistance of 50 $Ω$ and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100- $Ω$ resistance is

z e r o

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0.02 A

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0.04 A

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0.046 A

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Solution

The correct option is B $0.04 A$
Current cannot flow through the resistor $50 Ω$ in the middle because the diode is in reverse biased. Thus, it can be removed from the circuit and corresponding circuit is shown in figure.
Current through $100 Ω$ resistor is $I = \frac{6}{100 + 50} = 0.04 A$

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The circuit shown in the figure contains two diodes, each with a forward resistance of 50 Ω and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100-Ω resistance is

The correct option is B 0.04ACurrent cannot flow through the resistor 50Ω in the middle because the diode is in reverse biased. Thus, it can be removed from the circuit and corresponding circuit is shown in figure.Current through 100Ω resistor is I=6100+50=0.04A

The circuit shown in the figure contains two diodes, each with a forward resistance of 50 $Ω$ and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100- $Ω$ resistance is

The correct option is B $0.04 A$
Current cannot flow through the resistor $50 Ω$ in the middle because the diode is in reverse biased. Thus, it can be removed from the circuit and corresponding circuit is shown in figure.
Current through $100 Ω$ resistor is $I = \frac{6}{100 + 50} = 0.04 A$