QJEEM19Ph6453: JEE Main 2019 Physics - Class 12 JEEM19Ph6453 Solution: When the switch S, in the circuit shown, is closed, then the…

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.