Class 12 NEET Physics Current Electricity Questions Part 1

Resistance of series combination of 3Ω and 1Ω is R1=3+1=4Ω,R2=8Ω
Let i be the total current in the circuit.
Current through R1 is
i1=

i×R2

R1+R2

=

i×8

12

=

2i

3

Current through R2 is
i2=

i×R1

R1+R2

=

i×4

12

=

i

3

Power dissipated in 3‌Omega resistor isP1=i12×3
Power dissipated in 8Ω resistor is
P2=i22×8
∴

P1

P2

=

i12×3

i22×8

or,

P1

P2

=

(2i∕3)2×3

(i∕3)2×8

=

12

8

=

3

2

P1=

3

2

×P2=

3

2

×2=3 watt
∴ Power dissipated across 3‌Omega resistor is3 watt.