The circuit shown below contains two ideal diodes, each with a forward resistance of 50Ω. If the battery voltage is 6V, the current through the 100Ω resistance (in Amperes) is :
A
0.027
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B
0.030
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C
0.020
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D
0.036
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Solution
The correct option is C0.020 As D2 is reversed biased , so no current through 75Ω resistor.
now Req=150+50+100=300Ω So, required current I=VReq=6300 I=6300=0.02A