The resistance P, Q and R in the circuit have equal resistance.

The battery of negligible resistance, supplies a total power of 12W. What is the power dissipated by heating in resistor R -

Options

A. 2W
B. 4W
C. 3W
D. 6W

Answer

A

Explanation

Let the resistances of P, Q and R be r :
The total resistances across the battery is
`R_("Total")=(rxxr)/(r+r)+r=(3)/(2)r`
Current through P `I=sqrt(("power")/(R_("Total")))=sqrt((12)/((3r)/(2)))=sqrt((8)/(r))`
Current through R, `I'=(I)/(2)=sqrt((2)/(r))`
`:.` Power dissipated in R
`P_(R)=(I')^(2)r=(2)/(r)xxr=2=2`watt