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.)

Given Information:  

Headlight power = Ph = 30.5 W

Starter power = Ps = 2.75 kW = 2750 W

Voltage = V = 12 V

Required Information:  

Pseries = ?

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

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

emily ruth's annual premium is $1058.50

step-by-step explanation:

emily ruth is a female of 27 years age purchasing 20 year endowment insurance with face value $25000.

from the given table,

for females, the age of a male 3 years younger should be used.

so, we have to check for 24 year male for 20 year endowment.

it is 42.34 for every $1000 face value.

for a face value of $25000,

it is 25 * 42.34 = $1058.50

∴ emily ruth's annual premium is $1058.50