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In the circuit shown the cell is ideal (EMF 10 V), the coil has inductance L = 4 H and zero resistance, and F is a fuse of zero resistance that blows when the current through it reaches 5 A. The switch is closed at t = 0. With a pure inductor across an ideal cell the current rises linearly. After how long does the fuse blow?

1. after 2s
2. just after t = 0
3. after 5s
4. after 10s

Correct answer: after 2s

Solution

With zero resistance anywhere, the cell EMF appears entirely across the inductor: EMF = L di/dt, so di/dt = EMF/L = 10/4 = 2.5 A/s, constant. The current rises linearly: i(t) = (EMF/L) t = 2.5 t. It reaches the fuse limit 5 A when 2.5 t = 5, i.e. t = 2 s. The fuse blows after 2 s. (It does not blow just after t = 0 because the inductor opposes the sudden current rise; the current builds up gradually and linearly.)

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