How Much Energy Does a Lamp Dissipate with a 0.3 A Current and 6 V Power Supply?

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SUMMARY

A lamp dissipates 216 Joules of energy when a current of 0.3 A is passed through it for 2 minutes using a 6 V power supply. The calculations confirm that the charge (Q) is 36 Coulombs, derived from the equation I=Q/t, where I is the current and t is the time in seconds. The energy (E) is then calculated using the formula E=QV, resulting in E=216 J. This method effectively demonstrates the relationship between current, voltage, and energy dissipation in electrical circuits.

PREREQUISITES
  • Understanding of basic electrical concepts such as current, voltage, and charge.
  • Familiarity with the equations I=Q/t and E=QV.
  • Knowledge of units of measurement for electrical energy (Joules, Amperes, Volts).
  • Basic grasp of power calculations in electrical circuits.
NEXT STEPS
  • Study the relationship between power, current, and voltage using the formula P=IV.
  • Learn about energy dissipation in resistive circuits and the role of resistance.
  • Explore the concept of electrical energy consumption over time and its implications.
  • Investigate the efficiency of different types of lamps and their energy usage.
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Students in electrical engineering, hobbyists experimenting with circuits, and anyone interested in understanding energy dissipation in electrical devices.

blackout85
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A current of 0.3 A is passed through a lamp for 2 minutes using a 6 V power supply. The electrical energy dissipated by this lamp during the two minutes is:
a) 1.8 J
b) 12 J
c) 20J
d) 36 J
e) 216 J

If someone could just help me to start off that would be great.
The main equations I have to use are

I=Q/t
E= QV

(.3A)= (Q)/(120)
Q= 36

E= (36)(6V)
E=216 J

Am I right:confused:
 
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looks good, tho ordinarily the power/heat would first be calculated as I*V, then multiplied by 120
 

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