JEE Main 16 March 2021 Shift 1 Solved Paper

According to the figure given in question, all 3Ω resistances are in parallel combination. So, their equivalent resistance is
‌

1

R‌parallel ‌

=‌

1

3

+‌

1

3

+‌

1

3

‌

1

R‌parallel ‌

=‌

3

3

⇒‌‌R‌parallel ‌=1kΩ5‌mathrm‌k‌Omega and 1‌mathrm‌k‌Omega resistance are in series to the equivalent of all 3‌Omega resistances.
∴‌‌R‌net ‌=5+1+R‌parallel ‌...(i)
⇒‌‌R‌net ‌=(5+1+1)kΩ=7kΩ=7×103Ω
∴ The value of electric current flowing through 5kΩ resistor will be
I=‌

V

R‌net ‌

=‌

21

7×103

=3×10−3A
=3mA
Comparing with the given value in the question i.e., xmA, the value of x=3.