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Standard XII

Physics

Conductivity and Resistivity

The circuit s...

Question

The circuit shown in the figure contains two diodes $D_{1}$ and $D_{2}$ each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6V, the current through the 100 ohm resistance (in Amperes) is:

Zero

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0.02

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0.03

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0.036

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Solution

The correct option is B
0.02

$D_{1} \to F . B (f o r w a r d b i a s), D_{2} \to R . B (r e v e r s e b i a s)$

$I = \frac{V_{net}}{R_{n e t}} I = \frac{6}{300}$

$= 0.02 A$

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Similar questions

The circuit shown in the figure contains two diodes $D_{1}$ and $D_{2}$ 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 (in $A$ ) is:

Q. The circuit shown in the figure contains two diodes each with a forward resistance of

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The circuit shown in the figure contains two diodes D1 and D2 each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6V, the current through the 100 ohm resistance (in Amperes) is:

The correct option is B 0.02 D1→F.B(forward bias) , D2→R.B(reverse bias) I=VnetRnet I=6300 =0.02 A

The circuit shown in the figure contains two diodes $D_{1}$ and $D_{2}$ each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6V, the current through the 100 ohm resistance (in Amperes) is:

The correct option is B
0.02

$D_{1} \to F . B (f o r w a r d b i a s), D_{2} \to R . B (r e v e r s e b i a s)$

$I = \frac{V_{net}}{R_{n e t}} I = \frac{6}{300}$

$= 0.02 A$