The current flowing through 10 Omega resistor in the circuit shown in the figure is
Answer»
50 mA 20 mA 40 mA 40 mA
Solution :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 `10Omega` RESISTOR is `I=(2V)/(10Omega+15Omega)=0.08A=80mA`