Answer:
(a) 277.05 Ω
(b) 15.39 Ω
(c) 3.76 W
Explanation:
(a)
Applying,
P = V²/R.......................... Equation 1
Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string
Make R the subject of the equation
R = V²/P................... Equation 2
Given: V = 130, P = 61 W
Substitute into equation 2
R = 130²/61
R = 277.05 Ω
(b) The resistance of a single light is given as
R' = R/18 (since the light are connected in series and the are identical)
Where R' = Resistance of the single light.
R' = 277.05/18
R' = 15.39 Ω
(c)
Heat dissipated in a single light is given as
P' = I²R'..................... Equation 3
Where P' = heat dissipated in a single light, I = current flowing through each light.
We can calculate for I using
P = VI
make I the subject of the equation
I = P/V
I = 61/130
I = 0.469 A.
Also given: R' = 15.39 Ω
Substitute into equation 3
P' = 0.496²(15.39)
P' = 3.76 W
