JEE Main 2019 — Current Electricity Question with Solution

We start by applying Kirchhoff's current law at point C in the circuit:

$$i_1+i_2=i$$

where $i_1$ and $i_2$ are the currents flowing through resistors R₁ and R₂ respectively, and i is the current flowing through the battery and resistor R₃.

Next, we use Ohm's law to express $i_1$ and $i_2$ in terms of the voltages at points A, B, and C:

$$i_1=\frac{V_A-V_C}{R_1}=\frac{20-V_C}{2}$$

$$i_2=\frac{V_B-V_C}{R_2}=\frac{10-V_C}{4}$$

where $R_1$ and $R_2$ are the resistances of resistors R1 and R2 respectively.

Substituting these expressions into the first equation, we get:

$$\frac{20-V_C}{2}+\frac{10-V_C}{4}=i$$

We then use Ohm's law again to express i in terms of the voltages at points C and earth :

$$i=\frac{V_C-V}{R_3}=\frac{V_C-0}{2}=\frac{V_C}{2}$$

Substituting this expression into the previous equation, we get:

$$\frac{20-V_C}{2}+\frac{10-V_C}{4}=\frac{V_C}{2}$$

Simplifying this equation, we get:

$$40-2V_C+10-V_C=2V_C$$

$$50=5V_C$$

$$V_C=10\text{ V}$$

Finally, we use Ohm's law to find the value of the current i:

$$i=\frac{V_C}{2}=\frac{10}{2}=5\text{ A}$$

Therefore, the current flowing through the circuit is 5 A.