The current flowing through 10Ω resistor in the circuit shown in the figure is: |
50 mA 20 mA 40 mA 80 mA |
80 mA |
In the given circuit, junction diode $D_1$ is forward biased and will conduct whereas junction diode $D_2$ is reverse biased and will not conduct. Therefore current through 10Ω resistor is $I=\frac{2V}{10Ω+15Ω}=0.08A = 80 mA$ |