How Do You Calculate Potential Difference in Car Battery Circuits?

AI Thread Summary
To calculate potential difference in car battery circuits, the formula PD (V) = Energy Supplied / Charge that flows is used, yielding a result of 12V for a headlamp receiving 6,000,000 J from 500,000 C of charge. For a power supply of 60W and a current of 5.0 A, the potential difference is calculated as 120V. Additional calculations include determining current and energy for various components, such as a 4600 W electric motor requiring 20 A from a 230 V supply. The discussion highlights the importance of understanding the relationship between power, current, and voltage in practical applications. Overall, the calculations demonstrate the application of electrical principles in real-world scenarios.
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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?
 
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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.
 
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