Question

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.

Answer 1

The resistance of the lighter is 46.54 ohms and the current delivered by the battery is 0.26 A.

Step-by-step explanation:

To find the resistance of the lighter, we can use the formula P = V^2 / R, where P is the power, V is the voltage, and R is the resistance. By rearranging the formula, we can solve for R, which is R = V^2 / P. Plugging in the values, we have R = (12V)^2 / 37W = 46.54 ohms.

To find the current that the battery delivers to the lighter, we can use Ohm's Law, which states that I = V / R, where I is the current, V is the voltage, and R is the resistance. Plugging in the values, we have I = 12V / 46.54 ohms = 0.26 A.

Answer 2

(a) 3.89Ω

(b) 3.08A

Step-by-step explanation:

(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