A 25.00 V battery is used to supply current to a 21 k? resistor. Assume the volt

Question

A 25.00 V battery is used to supply current to a 21 k? resistor. Assume the voltage drop across any wires used for connections is negligible. (a) What is the current (in mA) through the resistor? X mA (b) What is the power dissipated by the resistor (in mw)i? (c) What is the power input from the battery (in mW), assuming all the electrical power is dissipated by the resistor? (d) What happens to the energy dissipated by the resistor? O The energy dissipated by the resistor is converted to electric potential energy. O The energy dissipated by the resistor is converted to elastic potential energy. The energy dissipated by the resistor is converted to heat. O The energy dissipated by the resistor is converted to gravitational potential energy.

Explanation / Answer

A.

Current through the resistor

i= V/R= 25/21= 1.19 mA

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B.

P= i^2* R= (1.19*10^-3)^2* 21000= 0.03 W

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C. P= 0.03 W

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Do comment in case any doubt.. goodluck

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