Calculate the electrical energy used by a 240 v light bulb with a resistance of 60 (omega) in 600 s.​

Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE = qV, where q is the charge moved and V is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is

P

=

P

E

t

=

q

V

t

.

Recognizing that current is I = q/t (note that Δt = t here), the expression for power becomes

P = IV

Electric power (P) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅V= 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power P = IV = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (1 kA ⋅V = 1 kW). To see the relationship of power to resistance, we combine Ohm’s law with P = IV. Substituting I = V/R gives P = (V/R)V=V2/R. Similarly, substituting V = IR gives P = I(IR) = I2R. Three expressions for electric power are listed together here for convenience:

P

=

IV

P

=

V

2

R

P

=

I

2

R

.

Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, P can be the power dissipated by a single device and not the total power in the circuit.) Different insights can be gained from the three different expressions for electric power. For example, P = V2/R implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in P = V2/R, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.

i guess it’s a and b