When the lights of an automobile are switched on, an ammeter in series with them reads 9.70 A and a voltmeter connected across them reads 12.20 V. When the electric starting motor is also turned on, the ammeter reading drops to 7.30 A and the lights dim somewhat. If the internal resistance of the battery is 51 mΩ and that of the ammeter is negligible, what is the emf of the battery?
B)What does the voltmeter read when the starting motor is turned on?
C)What is the current through the starting motor?
I have figured out part A but I am having troubles with part B and C. Help?
To solve parts B and C, we need to consider the changes that occur when the starting motor is turned on.
Let's calculate the new current through the lights when the starting motor is turned on.
1. Calculate the equivalent resistance of the lights when only the lights are switched on:
Using Ohm's Law, we have V = I * R.
Rearranging the equation, R = V/I.
Here, V = 12.20 V and I = 9.70 A.
Therefore, the equivalent resistance of the lights is R1 = 12.20 V / 9.70 A.
2. Calculate the new current when the starting motor is turned on:
The new current reading on the ammeter is 7.30 A.
To find the current through the lights when the starting motor is turned on, we need to subtract the current through the starting motor from the total current.
Thus, the current through the lights is I1 = 7.30 A - I2 (current through the starting motor).
3. Calculate the current through the starting motor (I2):
The drop in current from 9.70 A to 7.30 A occurs due to the internal resistance of the battery (51 mΩ).
To calculate the current through the starting motor, we can use Kirchhoff's laws and voltage division.
a) Calculate the drop in voltage across the internal resistance of the battery:
Using Ohm's Law, we have V = I * R.
Rearranging the equation, V_internal = I * R_internal.
Here, R_internal = 51 mΩ (or 0.051 Ω).
Therefore, the drop in voltage across the internal resistance is V_internal = (9.70 A - 7.30 A) * 0.051 Ω.
b) Calculate the voltage across the starting motor:
The total voltage of the battery is equal to the voltage across the lights plus the drop across the internal resistance, which means:
Total voltage = V_lights + V_internal.
Rearranging the equation, V_lights = Total voltage - V_internal.
c) Calculate the current through the starting motor:
Using Ohm's Law, we have V = I * R.
Rearranging the equation, I_starting motor = V_lights / R_starting motor.
Now we can calculate parts B and C.
B) What does the voltmeter read when the starting motor is turned on?
The voltmeter is connected across the lights, so it will show the voltage across the lights.
Therefore, the voltmeter reading when the starting motor is turned on is equal to V_lights calculated in step 3b.
C) What is the current through the starting motor?
The current through the starting motor is equal to I_starting motor calculated in step 3c.
Sure! Let's break down the problem step by step.
A) To find the electromotive force (emf) of the battery, we can use Ohm's Law. Ohm's Law states that the voltage across a resistor (V) is equal to the current (I) multiplied by the resistance (R). In this case, the resistor is the internal resistance of the battery.
We know that the voltmeter reads 12.20 V and the ammeter reads 9.70 A when only the lights are switched on. So, we can write the equation as:
12.20 V = 9.70 A × (R_batt + R_internal)
Since the source voltage (emf) is given by the sum of the voltage across the battery's internal resistance and the external resistance (lights), we can rearrange the equation to solve for the emf of the battery:
Emf = V - I × R_internal
Emf = 12.20 V - 9.70 A × 51 mΩ
Solving this equation, we get:
Emf = 12.20 V - 9.70 A × 0.051 Ω
Emf = 12.20 V - 0.4947 V
Emf ≈ 11.71 V
Therefore, the electromotive force of the battery is approximately 11.71 V.
Now let's move on to part B and part C.
B) When the starting motor is turned on, the ammeter reading drops to 7.30 A, indicating a decrease in current. To find the voltmeter reading, we need to determine the voltage across the lights while the motor is running.
We know that the total current flowing through the circuit is the sum of the currents through the lights (I_light) and the starting motor (I_motor):
I_total = I_light + I_motor
From our previous calculations, the total current (I_total) decreased from 9.70 A to 7.30 A. Since the ammeter is in series with the lights, we can conclude that the current through the lights decreased as well:
I_light = 9.70 A - 7.30 A = 2.40 A
Now, to find the voltmeter reading (V_light_motor), we can use Ohm's Law again. Rearranging the equation, we have:
V_light_motor = I_light × R_light
We don't have the value for the resistance of the lights (R_light), so we need to make an assumption here. Let's assume that the resistance of the lights remains constant. Therefore, the voltmeter reading when the starting motor is turned on can be found using the equation above.
C) The current through the starting motor (I_motor) can be found by subtracting the current through the lights (I_light) from the total current (I_total):
I_motor = I_total - I_light
I_motor = 7.30 A - 2.40 A
I_motor ≈ 4.90 A
Therefore, the current through the starting motor is approximately 4.90 A.
Keep in mind that the values obtained for part B and part C may vary depending on the assumptions made about the resistance of the lights.