QJEEM19Ph7711: JEE Main 2019 Physics - Class 12 JEEM19Ph7711 Solution: Determine the charge on the capacitor in the following circuit :

Assuming potential x and y at the top junctions A and B respectively and zero potential at lower junctions.

At the junction A applying KCL,

i₁ + i₂ + i₃ = 0

$$\frac{x-72}{6}+\frac{x}{4}+\frac{x-y}{2}=0$$

$$\Rightarrow$$ $$\frac{2x-144+6x-6y+3x}{12}=0$$

$$\Rightarrow$$ 11x - 6y = 144 ......(1)

At the junction B applying KCL,

i₄ + i₅ = 0

$$\frac{y-x}{2}+\frac{y}{10}=0$$

$$\Rightarrow$$ $$\frac{5y-5x+y}{10}$$ = 0

$$\Rightarrow$$ 6y = 5x ......(2)

Solving (1) and (2)

11x - 5x = 144

$$\Rightarrow$$ x = 24 V

$$\therefore$$ y = 20 V

The charge on the capacitor, q = CV

$$\Rightarrow$$ q = 10$$\mu$$F × 20 = 200 $$\mu$$C