QJEEM21Pha4a7: JEE Main 2021 Physics - Class 12 JEEM21Pha4a7 Solution: The circuit contains two diodes each with a forward resistance…

Given, forward resistance, R₁ = 50 $$\mathrm{\Omega }$$

Reverse resistance, R₂ = infinity

Battery voltage = 6V

According to circuit diagram,

In this case, diode D₁ is forward biased, whereas diode D₂ is reverse biased.

So, D₂ will act as open circuit.

$$6-50I-130I-120I=0$$

$$\Rightarrow 6=300I$$

$$\Rightarrow I=\frac{6}{300}=\frac{1}{50}$$

$$=\frac{2}{100}=0.02$$ A = 20 mA

Hence, current through 120 $$\mathrm{\Omega }$$ = 20 mA