CGPET 2018 Solved Paper

In the circuit shown in the figure below.
So, voltage across 50Ω and 30Ω resistor,
V1=I1×50=0.1×50=5V
V2=I1×30=0.1×30=3V
∵ Three branches are connected in parallel combination, therefore voltage across each branch will be equal.
∴V1+V2=V3
⇒ 5+3=V3
⇒ V3=8V
Applying KCL at point B,
I2=I1+I3=0.1+

2

5

=0.5A
Hence, V1+V2=12−I2R
⇒R=

12−8

0.5

=8Ω