Given Information:
Headlight power = Ph = 30.5 W
Starter power = Ps = 2.75 kW = 2750 W
Voltage = V = 12 V
Required Information:
Pseries = ?
Answer:
Pseries = 30.2 W
Explanation:
First we have to find the resistance of headlight and starter
As we know power is given by
P = V²/R
For headlight:
Ph = V²/Rh
30.5 = 12²/Rh
Rh = 12²/30.5
Rh = 4.72 Ω
For starter:
Ps = V²/Rs
2750 = 12²/Rs
Rs = 12²/2750
Rh = 0.0523 Ω
In a series connection, the equivalent resistance will be
Req = Rh + Rs
Req = 4.72 + 0.0523
Req = 4.772 Ω
Therefore, the power consumed by headlight and the starter is
Pseries = V²/Req
Pseries = 12²/4.772
Pseries = 30.18
Pseries = 30.2 W
Answer:
Headlight, P = 29.7W and starter P = 0.330W.
Explanation:
Given V = 12V
P = I×V = V/R ×V = V²/R
So the resistance for each element is given by R = V²/P
So for the head light,
P = 30.5W,
R1 = 12²/30.5 = 4.72Ω
For the starter P = 2.75kW = 2750W
R2 = 12²/2750 = 0.0524Ω
Now in series connection the same current I flows through them.
So Req = R1 + R2 = 4.72 + 0.0524 = 4.7724Ω
I = V/Req = 12/4.7724 = 2.51A
So the power consumed by:
The headlight is P = I²R1 = 2.51²×4.72 = 29.7W.
The Starter P = I²×R2 = 2.51²×0.0524 = 0.330W
