If power dissipated in the `9 Omega` resistor in the resistor shown is `36 W`, the potential difference across the `2 Omega` resistor is

Options

A. 4 volt
B. 8 volt
C. 10 volt
D. 2 volt

Answer

c

Explanation

As, `P = I^(2)R or I = sqrt(P//R)`
`:.` Current through `9 Omega` resistor given below is
`I_(1) = sqrt((36)/(9)) = 2A`

As resistors `9 Omega` and `6 Omega` are connected in parallel therefore potential difference
`V_(A) - V_(B) = 9I_(1) = or I_(2) = 9I_(1)//6 = (9xx2)/(6) = 3A`
Current drawn from the battery `= I_(1) + I_(2) = 2+ 3 = 5A`
Pot diff `2 Omega` resistor `= (5A)( 2 Omega) = 10V`