Answer:
1.45 Ω
Explanation:
Resistance of each bulb
P = V²/R.............. Equation 1
Where P = power, V = Voltage, R = Resistance.
R = V²/P............ Equation 2
Given: V = 12 V, P = 4 W
R = 12²/4
R = 36 Ω
Since the bulbs are connected in parallel,
Total resistant is given as
Rt = RR'/(R+R')
Where Rt = Total resistance
Rt = (36×36)/(36+36)
Rt = 18 Ω
The current flowing through the Circuit is given as
I = V/R................... Equation 3
Where I = current.
Note: The Voltage delivered to the bulb is 11.1 V.
Given: V = 11.1 V, R = 18 Ω
Substitute into equation 3
I = 11.1/18
I = 0.62 A.
Finally,
Lost Voltage = internal resistance×current
E-V = r×I................... Equation 4
Where E = E.M.F of the battery, r = internal resistance.
Make r the subject of the equation,
r = (E-V)/I................ Equation 5
Given: E = 12.0 V, V = 11.1 V, I = 0.62 A.
Substitute into equation 5
r = (12-11.1)/0.62
r = 1.45 Ω
Hence the internal resistance = 1.45 Ω
