Understanding Electrical Circuits and Power: Questions 10&11

In summary: As for 10d), your answer is also correct. Remember, power is the rate at which work is done, so the lower value makes sense for a longer period of time. For Q11), you could use the formula P=W/t to calculate the power needed for the car to go up the hill at a speed of 90km/h for an hour. This could help you determine the work done and the GPE (gravitational potential energy) involved in this process.
  • #1
C0balt
7
0

Homework Statement


Questions 10&11 in pictures.

Homework Equations


P=IV
W=VIt
W=Pt

The Attempt at a Solution


For 10a) P=IV
P=70*14= 980W.
10b) P=IV
1500=12*I
I=125A (which is massive so probably wrong)
10c)P=IV
60=I*12
I=5A
The 8A fuse because the current flowing through each headlamp should be 5A. Anymore than that could be dangerous.
10d) W=VIt
W=12*1*62
W=744J (which is quite low??)
10e)
W=Pt
It just doesn't give me the right answer as the sidelights appear to use more than the whole battery capacity if left on for 12 hours, which makes me believe I've done the whole of Q10 wrong.
Q11)
Maybe you could work out the work done / GPE then you could use P=W/t to work out how much power the car needs to go up the hill? (Using an hour as t from the 90km/h?)

I would greatly appreciate any help.
 

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  • #2
C0balt said:

Homework Statement


Questions 10&11 in pictures.

Homework Equations


P=IV
W=VIt
W=Pt

The Attempt at a Solution


For 10a) P=IV
P=70*14= 980W.
10b) P=IV
1500=12*I
I=125A (which is massive so probably wrong)
10c)P=IV
60=I*12
I=5A
The 8A fuse because the current flowing through each headlamp should be 5A. Anymore than that could be dangerous.
10d) W=VIt
W=12*1*62
W=744J (which is quite low??)
10e)
W=Pt
It just doesn't give me the right answer as the sidelights appear to use more than the whole battery capacity if left on for 12 hours, which makes me believe I've done the whole of Q10 wrong.
Q11)
Maybe you could work out the work done / GPE then you could use P=W/t to work out how much power the car needs to go up the hill? (Using an hour as t from the 90km/h?)

I would greatly appreciate any help.

Your answer for 10b) is correct. Starter motors take a large amount of current to turn the engine over. They do so for only a short time, however, lest the battery be drained completely.
 

1. What is an electrical circuit?

An electrical circuit is a closed loop through which electrons can flow. It consists of a power source, such as a battery, wires or conductors to carry the current, and a load, such as a light bulb, that converts the electrical energy into another form of energy.

2. What is the difference between AC and DC circuits?

AC (alternating current) circuits have a constantly changing direction of flow for the electrons, while DC (direct current) circuits have a constant flow of electrons in one direction. AC is typically used for large-scale power distribution, while DC is used for smaller, portable devices.

3. How is power calculated in an electrical circuit?

Power in an electrical circuit is calculated using the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amps. This formula can also be rearranged to find voltage (V = P/I) or current (I = P/V).

4. What is the purpose of a circuit breaker or fuse?

Circuit breakers and fuses are safety devices that protect an electrical circuit from excess current. When the current exceeds a certain limit, the circuit breaker or fuse will trip and break the circuit, preventing overheating and potential fire hazards.

5. How does resistance affect an electrical circuit?

Resistance is the measure of how much a material or component impedes the flow of electricity. In an electrical circuit, resistance can affect the amount of current and voltage in the circuit. Higher resistance means less current can flow, while lower resistance allows for more current to flow.

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