The circuit shown below contains two ideal diodes, each with a forward resistance of $50 \Omega$. If the…
The circuit shown below contains two ideal diodes, each with a forward resistance of $50 \Omega$. If the battery voltage is $6 \mathrm{~V},$ the current through the $100 \Omega$ resistance (in Amperes) is:
0.036
0.02
0.027
0.03
Solution
As $D_{2}$ is reversed biased, so no current through $75 \Omega$ resistor.
now $R_{e q}=150+50+100$
$=300 \Omega$
So, required current $I=\frac{\text { BatteryVoltage }}{300}$ $\mathrm{I}=\frac{6}{300}=0.02$
