In above shown figure, a circuit has resistor R and battery of emf
ε. If current I is flowing through circuit
then power dissipated as heat is P. The circuit is changed by doubling
the emf ε of the battery while R is kept
constant. After the change, the power dissipated in R is:
Given - resistance in circuit R1=R , emf ε1=ε and power P1=P
power of circuit
P=ε2/R ............eq(1)
when emf is doubled and resistance is kept constant
R2=R , ε2=2ε , P2=?
P2=((2ε)2)/R..............eq(2)
dividing eq(2) by eq(1)
P2=4P