Class 12 NEET Physics Current Electricity Questions Part 1

Current flows through the 9 Ω resistor is
I12=

36

9

=4  (As P=I2R)
I1=2A
As the resistors 9 Ω and 6 Ω are connected in parallel, therefore potential difference across them is same.
∴9I1=6I2;I2=

9×2

6

=3A
Current drawn from the battery is
I=I1+I2=(2+3)A=5A
The potential difference across the 2 Ω resistor is
=(5A)(2Ω)=10V