Energy dissipated as heat during a time interval over resistor with constant pot. dif

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SUMMARY

The energy dissipated as heat by a 1.5 kΩ resistor with a constant 20 V potential difference over a 2-minute interval can be calculated using the formula for electrical power. The relevant equations include \( P = \frac{V^2}{R} \) for power and \( E = P \times t \) for energy, where \( P \) is power, \( V \) is voltage, \( R \) is resistance, and \( t \) is time. Substituting the values, the power is 0.267 W, leading to an energy dissipation of 32.04 J over the specified time period.

PREREQUISITES
  • Understanding of Ohm's Law and electrical power calculations
  • Familiarity with the formula \( P = \frac{V^2}{R} \)
  • Knowledge of energy calculations using \( E = P \times t \)
  • Basic concepts of resistors and voltage in electrical circuits
NEXT STEPS
  • Study the derivation and application of Ohm's Law in circuit analysis
  • Learn about the relationship between power, voltage, and resistance in electrical systems
  • Explore energy dissipation in resistive components in detail
  • Investigate the effects of varying voltage and resistance on energy dissipation
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit design or analysis will benefit from this discussion.

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Homework Statement


How much energy is dissipated as heat over 2 minute time interval by a 1.5 k(ohm) resistor which has a constant 20 V potential difference across its leads?


Homework Equations


deltaV=-Ed


The Attempt at a Solution


In my notes this is the only equation that seems like it might work here. I assume I probably have to use some other equations to figure out the d
 
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deltaV=-Ed
The E in this formula is electric field, which you don't have in this problem.
You need a formula for electric energy or a more general one relating power with energy along with a formula relating power with voltage and resistance.
 

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