A 4-W night-light is plugged into a 120-V circuit and oper - ates continuously for 1 year. Find the following: (a) the current it draws, (b) the resistance of its filament, (c) the energy consumed in a year. (d) Then show that for a utility rate of 15¢/kWh the cost for a year's operation is $5.25.

(a) 0.033 A.

(b) 3600 Ω

(c) 35.04 kWh

Explanation:

(a)

Electric power = Voltage * current.

P = VI ... Equation 1

Where P = Power, V = voltage, I = current.

Making I the subject of the equation,

I = P/V ... Equation 2.

Given: P = 4 W, V = 120 V.

Substitute into equation 2

I = 4/120

I = 0.033 A.

(b)

From ohm's law,

V = IR

R = V/I ... Equation 3

Where R = Resistance Filament.

Given: V = 120 V, I = 0.033 A,

Substitute into 3,

R = 120/0.033

R = 3600 ohm's.

(c)

Power = Energy/time.

P = E/t

E = P*t ... Equation 4

Where E = Energy consumed, t = time.

Given: P = 4 W, t = 1 year = 1*365*24 = 8760 hours.

Substitute into equation 4

E = 4*8760

E = 35040 Wh

E = 35.04 kWh

(d)

If 1 kWh = 15 ¢

Then 3.504 kWh = (15*35.04) ¢

= 525.6 ¢

If 100¢ = $1,

Then 525.6¢ = 5.256$.

Hence the cost for a year's operation = 5.25$