Your car's 30.5 W headlight and 2.75 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would one headlight and the starter consume if connected in series to a 12.0 V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices. Answer to the nearest 0.1 W.)

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

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