Question

An automobile battery has an emf of 12.6 V and an internal resistance of 0.080 0 V. The headlights together have an equivalent resistance of 5.00 V (assumed constant). What is the potential difference across the headlight bulbs (a) when they are the only load on the battery and (b) when the starter motor is operated, with 35.0 A of current in the motor?

Answer 1

Answer:

The question is incomplete, below is the complete question

"An automobile battery has an emf of 12.6 V and an internal resistance of 0.080 0 Ohm. The headlights together have an equivalent resistance of 5.00 Ohm (assumed constant). What is the potential difference across the headlight bulbs (a) when they are the only load on the battery and (b) when the starter motor is operated, with 35.0 A of current in the motor?"

Answers:

a. 12.4V

b. 9.6V

Explanation:

we can assume the arrangement of the resistors to be in series and from the rules of combining Resistor we can determine the equivalent resistance

$R_{eq}=R_{1}+R_{2}\\R_{eq}=0.080+5.0\\R_{eq}=5.080 ohms$

a. to determine the voltage across the headlight of the lamp, we use the voltage divider rule

$V_{lamp}=V\frac{R_{lamp}}{R_{eq}}\\V_{lamp}=12.6\frac{5}{5.080}\\V_{lamp}=12.4V$

b. The current current in the headlamp is

$I_{lamp}=V_{lamp}/R_{lamp}\\I_{lamp}=12.4/5\\I_{lamp}=2.48A$

The total current in the system when the motor is operated with a starting current of 35A is

2.48+35=37.48A

Next we get the voltage drop across the internal resistance at this current rating

$V_{r}=37.48*0.80\\V_{r}=3.0272A$

Hence the current in the Headlamp is now

$V_{lamp}=V-V_{r}\\V_{lamp}=12.6-3.0272\\V_{lamp}=9.6V$