The circuit shown below contains two ideal diodes- each with a forward resistance of 50 - If the battery voltage is 6 V- the current through the 100 - resistance in Amperes is -

As D_2 is reversed biased , so no current through 75 Ω resistor. now R_eq=150+50+100 nbsp; =300 Ω So, required current I=VR_eq=6300 I=6300=0.02 A Hence, (B) ...

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Standard XII

Physics

Power

The circuit s...

Question

The circuit shown below contains two ideal diodes, each with a forward resistance of $50 Ω$ . If the battery voltage is $6 V$ , the current through the $100 Ω$ resistance (in Amperes) is :

0.027

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0.030

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0.020

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0.036

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Solution

The correct option is C $0.020$
As $D_{2}$ is reversed biased , so no current through $75 Ω$ resistor.
now $R_{e q} = 150 + 50 + 100$ $= 300 Ω$
$So, required current I = \frac{V}{R_{e q}} = \frac{6}{300}$
$I = \frac{6}{300} = 0.02 A$

Hence, $(B)$ is the correct answer.

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The circuit shown below contains two ideal diodes, each with a forward resistance of 50 Ω. If the battery voltage is 6 V, the current through the 100 Ω resistance (in Amperes) is :

The correct option is C 0.020As D2 is reversed biased , so no current through 75 Ω resistor. now Req=150+50+100 =300 Ω So, required current I=VReq=6300 I=6300=0.02 A Hence, (B) is the correct answer.

The circuit shown below contains two ideal diodes, each with a forward resistance of $50 Ω$ . If the battery voltage is $6 V$ , the current through the $100 Ω$ resistance (in Amperes) is :

The correct option is C $0.020$
As $D_{2}$ is reversed biased , so no current through $75 Ω$ resistor.
now $R_{e q} = 150 + 50 + 100$ $= 300 Ω$
$So, required current I = \frac{V}{R_{e q}} = \frac{6}{300}$
$I = \frac{6}{300} = 0.02 A$

Hence, $(B)$ is the correct answer.