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Fantom [35]
3 years ago
5

A series circuit has a voltage supply of 12 V and a resistor of 2.4 kΩ. How much power is dissipated by the resistor?

Physics
1 answer:
adell [148]3 years ago
4 0

Answer:

(B) 0.06W

Explanation:

power absorbed by the resistor is given by \frac{V^2}{R}=I^2R

Where V = voltage

           R= resistance

            I =  current through the circuit

we have given V =12 Volt and resistance =2.4 K\Omega

current I=\frac{V}{R}=\frac{12}{2.4\times 1000}=5\times 10^{-3}A=5mA

power using voltage and resistance equation

p=\frac{12^2}{2.4\times 1000}=60mW =0.06W

using current equation P=I^2R=5^2\times 12=60mW= 0.06W

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Answer:

Explanation:

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Hope this helps.
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