How much power is produced if a voltage of #4 V# is applied to a circuit with a resistance of #2 Omega#?

1 Answer

#8 J/s or 8 W#

Explanation:

#P = IV = I^2 R=V^2/R#

What are given:
#V=4 V#
#R = 2\Omega#

Therefore,
#P = V^2/R =(4 (kg\cdotm^2\cdot s^-3\cdotA^-1))^2/(2(kg\cdot m^2 \cdot s^-3 \cdot A^-2))=8 (kg\cdot m^2 \cdot s^-3 )=8 J/s#

For the determination of units,
#V = Ed= (F/q_0)(d) = (mad)/(q_0) => (kg *m/s^2 \cdot m)/C => (kg *m/s^2 \cdot m)/(As)#

And

#\Omega=V/A = (kg *m/s^2 \cdot m)/(A^2s)#

*Take note that the equation #P = I^2 R=V^2/R# can be played using #P=IV# using the different fundamental equations of electricity.

#V = Ed =(F/q_0)(d) #
#\Omega=V/A#