Question

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0790 Ω. The headlights together have an equivalent resistance of 4.00 Ω (assumed constant). (a) What is the potential difference across the headlight bulbs when they are the only load on the battery? (Enter your answer to at least two decimal places.) V (b) What is the potential difference across the headlight bulbs when the starter motor is operated, with 35.0 A of current in the motor? (Enter your answer to at least two decimal places.)

Answer 1

The potential difference across the headlight bulbs when they are the only load on the battery is approximately 0.108V. When the starter motor is operated with a current of 35.0A, the potential difference across the headlight bulbs is approximately 7.044V.

Step-by-step explanation:

(a) To determine the potential difference across the headlight bulbs when they are the only load on the battery, we can use the formula V = emf - (I * R), where V is the potential difference, emf is the electromotive force of the battery, I is the current, and R is the resistance. Since the headlights have an equivalent resistance of 4.00 Ω, and no other load is present, the current flowing through them is I = emf / (internal resistance + resistance of headlights). Plugging in the given values, we have I = 12.6V / (0.0790Ω + 4.00Ω) ≈ 3.123A. Now we can calculate the potential difference across the headlight bulbs using V = 12.6V - (3.123A * 4.00Ω) ≈ 0.108V.

(b) When the starter motor is operated with a current of 35.0A, the potential difference across the headlight bulbs can be found using the same formula. In this case, the current flowing through the headlight bulbs is I = emf / (internal resistance + resistance of headlights). Plugging in the given values, we have I = 12.6V / (0.0790Ω + 4.00Ω + 4.00Ω) ≈ 1.083A. Now we can calculate the potential difference across the headlight bulbs using V = 12.6V - (1.083A * 4.00Ω) ≈ 7.044V.