How Do You Calculate Potential Difference in Car Battery Circuits?

[plaly]

hi, i got this hwk today and I've answered the questions i can do, can you check the answers i have done, and also help me with the ones i can't do.
thank you so much!
x

1. A car battery can supply 6 000 000 J to a car headlamp while delivering 500 000 C of charge through the headlamp. Calculate the potential difference across the headlamp.
PD (V) = Energy Supplied / Charge that flows
= 600 0000J / 500 000 C
= 12V

2. A car battery supplies a power of 60W to a car headlamp using a current of 5.0 A. Calculate the potential difference across the headlamp.
PD (V) = Power Supplied / Current
= 60 W / 5.0 A
= 120 V

3. If questions 1 and 2 apply to the same battery, under the same conditions of use, for how long was the battery in use?

4. A 120 W bulb for a projector could be manufactured so that it either operates on 240 V mains or on a 20 V transformer. How much current would be supplied by each supply?.
240 V mains current supplied -> 120 W / 240 V = 0.5 A
20 V transformer supplied -> 120 W / 20 V = 6 A

5a) A 4600 W electric motor is to operate from a 230 V mains supply. What current will be required?
Power / P.D = Current -> 4600 W / 230 V = 20 A

b) Why might more than this current be needed in practice to obtain this output power?

6. A current of 3.2 mA passes through a circuit component for 300 s. The PD across the component is 12 V. How much energy is supplied to the component?
Charge = Current x Time
= 0.0032 x 300
= 0.96 C
PD x Charge = Energy -> 12 x 0.96 = 11.52 W

7. During one use of a hairdrier the following quantities are measured. Determine the unknown quantities. Measured quanitites: PD = 240 V, power = 1200 W, time = 600 s. Unknown quantities: current, charge, energy.
Current = Power supplied / PD = 1200 W / 240 V = 5A
Charge = Current x Time = 5 A x 600 s = 3000 C
Energy = Power x Time = 1200 W x 600 s = 720000 W

b) Why might more than this current be needed in practice to obtain this output power?

 

[TVP45]

In question 2, you seem to have misplaced a decimal point.

You seem to be generally doing things correctly, but some of these are rather strange questions.

 

[m2003]

In practice, the actual current required may be slightly higher due to factors such as resistance in the circuit and energy losses. This means that the actual current needed to obtain the desired output power may be slightly higher than the calculated value. Additionally, the power rating of the hair dryer may not be completely accurate, leading to a slightly higher current needed to achieve the desired output power.