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The flux in a closed circuit of resistance 20ohm varies with time according to the equation phi = 6t^2-5t+1. What is the induced current at time t= 0.25 second? |
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Answer» SOLUTION :`phi = 6t^2-5t+1e = -(dphi)/DT = -[12t-5] ` at time t = 0.25s, E = `-[12xx0.25-5] = 2V ` `THEREFORE i = e/R = 2/20 = 0.1A` |
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