If the current in a resistor doubles while resistance and time remain constant, how does the heat energy q change?

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Multiple Choice

If the current in a resistor doubles while resistance and time remain constant, how does the heat energy q change?

Explanation:
The important relationship here is that heat energy generated in a resistor depends on the square of the current when resistance and time are fixed. Heat energy, q, from a resistor over a time t is q = I^2 R t. With resistance and time constant, q grows proportional to I^2. If the current doubles, substitute 2I into the equation: q' = (2I)^2 R t = 4 I^2 R t = 4 q. So the heat energy becomes four times larger. This also matches the idea that power dissipated is P = I^2 R, and over the same time, energy q = P t, so doubling I makes power—and thus the total heat—quadruple.

The important relationship here is that heat energy generated in a resistor depends on the square of the current when resistance and time are fixed. Heat energy, q, from a resistor over a time t is q = I^2 R t. With resistance and time constant, q grows proportional to I^2. If the current doubles, substitute 2I into the equation: q' = (2I)^2 R t = 4 I^2 R t = 4 q. So the heat energy becomes four times larger. This also matches the idea that power dissipated is P = I^2 R, and over the same time, energy q = P t, so doubling I makes power—and thus the total heat—quadruple.

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