A cigarette lighter in a car is a resistor that, when activated, is connected across the 12-V battery. Suppose a lighter dissipates 37 W of power. Find (a) the resistance of the lighter and (b) the current that the battery delivers to the lighter.

(a)  3.89Ω

(b) 3.08A

(a) Power(P) is related to voltage(V) and resistance(R) as follows;

P = V² / R     (i)

From the question; the following are given;

P = Power dissipated by the lighter = 37W

V = Voltage of the battery = 12V

Substitute these values into equation (i) as follows;

37 = 12² / R

37 = 144 / R

Solve for R;

R = 144 / 37

R = 3.89Ω

Therefore the resistance of the lighter is 3.89Ω

(b) The power(P) is also related to current (I) and voltage (V) as follows;

P = I x V

Substituting the values of P = 37W and V = 12V gives;

37 = I x 12

Solve for I;

I = 37 / 12

I = 3.08A

Therefore, the current that the battery delivers to the lighter is 3.08A

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

i think the answer is a and b

i wish you luck