QJEEM19Ph9881: JEE Main 2019 Physics - Class 12 JEEM19Ph9881 Solution: In the given circuit the the internal resistance of the 18 V…

Voltage accross resistance R₄ = 5 V

$$\therefore$$   IR₄ = 5 V

$$\Rightarrow$$   500 $$\times$$ I = 5

$$\Rightarrow$$   I = $$\frac{1}{100}$$ A

$$\therefore$$   Voltage across resistor R₃ = $$\frac{1}{100}\left(100\right)$$ = 1 A

$$\therefore$$   Total drop in resistance R₃ and R₄ = 5 + 1 = 6V

So, voltage accross R₂ resistance is also 6V as R₃, R₄ and R₂ are in parallel

$$\therefore$$   Voltage accross R₁ resistor R₁ resistor = 18 $$-$$ 6 = 12 V

$$\therefore$$   Current through R₁ resistor = $$\frac{12}{400}$$ = $$\frac{3}{100}$$ A

$$\therefore$$   Current through R₂ resistor

= $$\frac{3}{100}-\frac{1}{100}$$

= $$\frac{2}{100}$$ A

$$\therefore$$   $$\left(\frac{2}{100}\right)$$ R₂ = 6

$$\Rightarrow$$   R₂ = 300 $$\mathrm{\Omega }$$