A sinusoidal voltage \(V(t) = 100\sin(500t)\) is applied across a pure inductance of \(L = 0.02\mathrm{H}\). The current through the coil is:
Show Hint
- \(10\cos(500t)\)
- \(-10\cos(500t)\)
- \(10\sin(500t)\)
- \(-10\sin(500t)\)
The Correct Option is B
Solution and Explanation
Peak current \(I_0 = \dfrac{V_0}{X_L} = \dfrac{100}{10} = 10\ \mathrm{A}\).
In a pure inductor the current lags the voltage by \(\pi/2\): \(I = 10\sin\left(500t - \dfrac\pi2\right) = -10\cos(500t)\).
Answer: (2) \(-10\cos(500t)\)
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