In a purely resistive circuit, if the applied voltage is doubled, what happens to power?

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Enhance your understanding of the fundamentals of electricity with the AMT General Exam. Study with multiple-choice questions crafted to improve your knowledge and confidence. Prepare effectively for your success!

Multiple Choice

In a purely resistive circuit, if the applied voltage is doubled, what happens to power?

Explanation:
In a purely resistive circuit, the power dissipated by a resistor is proportional to the square of the voltage across it (P = V^2 / R) and also equals VI with I = V/R. Doubling the voltage doubles the current as well, since I = V/R. Then P = VI becomes (2V)(2I) = 4VI, which is four times the original power. Using P = V^2 / R, substituting 2V gives (2V)^2 / R = 4V^2 / R, again four times the original. So the power quadruples when the applied voltage is doubled.

In a purely resistive circuit, the power dissipated by a resistor is proportional to the square of the voltage across it (P = V^2 / R) and also equals VI with I = V/R. Doubling the voltage doubles the current as well, since I = V/R. Then P = VI becomes (2V)(2I) = 4VI, which is four times the original power. Using P = V^2 / R, substituting 2V gives (2V)^2 / R = 4V^2 / R, again four times the original. So the power quadruples when the applied voltage is doubled.

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