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A series RL circuit with a constant emf E, inductance L, and resistance R is closed. The current rises over time according to curve 1. When the circuit is closed again after changing one parameter (either L or R), the current follows curve 2, which rises more steeply and reaches a higher steady-state value than curve 1. Which parameter was changed and in what direction?

1. L is increased
2. L is decreased
3. R is increased
4. R is decreased

Correct answer: R is decreased

Solution

The steady-state current I = E/R increases only if R decreases. Decreasing R also shortens the time constant tau = L/R, so the current rises faster. Both observations (higher final value and steeper initial rise) are consistent with R being decreased.

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