UPSEE 2018 Solved Paper

Given circuit belowResistance between B and C on the right hand side of the circuit
=

10×10

10+10

=

100

20

=5Ω
because 1Ω+4Ω+5Ω=10Ω are in parallel with 10Ω.
Resistance between A and D on the right hand side of the circuit
=

10×10

10+10

=

100

20

=5Ω
because 5Ω (equivalent resistance of 1Ω,4Ω,5Ω and 10Ω ) is in series with 3Ω and 2Ω.
Hence, resistance between A and D will be 5Ω.
Equivalent resistance of the circuit,
R=5+2+5=12Ω
Current drawn from the battery,
I=

12V

12Ω

=1A
At the junction A, the current of 1A is divided equally between the 10Ω resistance and the remaining circuit of the 10Ω resistance.
At the junction B, the current of 0.5A is divided equally between the 10Ω resistor and remaining circuit of resistance 10Ω.
∴ Current through the 4‌Omega resistor =0.25A.