How Do You Calculate Potential Difference in Car Battery Circuits?

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SUMMARY

This discussion focuses on calculating potential difference (PD) in car battery circuits using specific formulas. The potential difference across a headlamp powered by a car battery is calculated as 12V when 6,000,000 J of energy is supplied through 500,000 C of charge. Another calculation shows a PD of 120V when a 60W power is supplied at a current of 5.0 A. Additional calculations involve determining current and energy supplied across various components, including a 4600 W electric motor and a hairdryer. The discussion emphasizes the importance of accurate calculations in electrical circuits.

PREREQUISITES
  • Understanding of Ohm's Law and basic electrical formulas
  • Familiarity with units of measurement: volts (V), watts (W), and coulombs (C)
  • Knowledge of power calculations in electrical circuits
  • Ability to manipulate equations for current, voltage, and energy
NEXT STEPS
  • Study the relationship between power, voltage, and current using Ohm's Law
  • Learn how to calculate energy consumption in electrical devices
  • Explore the implications of current ratings in practical applications
  • Investigate the effects of resistance on potential difference in circuits
USEFUL FOR

Students studying physics, electrical engineers, and anyone involved in automotive electrical systems will benefit from this discussion on calculating potential difference in car battery circuits.

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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?
 
Last edited:
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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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