Electromotive Force: Calculate Charge, E.M.F & P.D.

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SUMMARY

The discussion focuses on calculating the charge, electromotive force (E.M.F.), potential difference (P.D.), and lost volts for an Atorch battery supplying a current of 0.25A for 4400 seconds. The charge that flows is calculated as 1100 Coulombs using the formula Q=IT. The E.M.F. of the battery is determined to be 4.73V by dividing the total energy transferred (5200J) by the charge (1100C). The potential difference across the bulb can be calculated using the formula V=W/Q, where W is the energy transferred to the bulb (4800J). The lost volts can be found by subtracting the bulb voltage from the E.M.F.

PREREQUISITES
  • Understanding of electrical concepts such as charge, current, and voltage.
  • Familiarity with the formula Q=IT for calculating charge.
  • Knowledge of energy transfer in electrical circuits.
  • Ability to apply the formula V=W/Q for calculating potential difference.
NEXT STEPS
  • Learn how to calculate power in electrical circuits using the formula P=W/t.
  • Study the concept of internal resistance in batteries and its effect on E.M.F.
  • Explore the relationship between energy, charge, and voltage in electrical systems.
  • Investigate practical applications of E.M.F. and P.D. in real-world circuits.
USEFUL FOR

Students studying physics, particularly those focusing on electricity and circuits, as well as educators looking for practical examples of E.M.F. and P.D. calculations.

Xazdrubal
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Homework Statement



Atorch battery transfers 5200J of chemical energy into electrical energy while supplying a current of 0.25A for 4400 s. The bulb transfers 4800 j of electrical energy into heat and light during this time. Calculate
a)the charge that flows
b)the e.m.f. of the battery
c) the p.d. across the bulb
d)the lost volts

Homework Equations



Q=IT


The Attempt at a Solution


a) 0.25 x 4400 = 1100 C
b)5200/1100 = 4.73V
C) V=W/Q

NOT SURE WHAT TO DO FOR C / D /
 
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Xazdrubal said:

Homework Statement



Atorch battery transfers 5200J of chemical energy into electrical energy while supplying a current of 0.25A for 4400 s. The bulb transfers 4800 j of electrical energy into heat and light during this time. Calculate
a)the charge that flows
b)the e.m.f. of the battery
c) the p.d. across the bulb
d)the lost volts

Homework Equations



Q=IT


The Attempt at a Solution


a) 0.25 x 4400 = 1100 C
b)5200/1100 = 4.73V
C) V=W/Q

NOT SURE WHAT TO DO FOR C / D /

Welcome to the PF.

What does "p.d. across the bulb" mean? Power dissipated?

If so, then power is just total energy divided by time, right?

Once you have the bulb power, calculate the bulb voltage, and the difference between that and the battery source EMF is what is lost across the battery's internal resistance.
 

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