Ncert Exercises & Solutions

A (current versus time) graph of the current passing through a solenoid is shown in the figure. For which time is the back electromotive force (u) a maximum. If the back emf at t=3 s is e, find the back emf at t=7 s, 15s and 40 s. OA, AB and BC are straight line segments.

Hint: The back emf depends on the rate of change of the current.

Step 1: The back emf (u) in the solenoid is a maximum when there is a maximum rate of change of current. This occurs is in part AB of the graph. So maximum back emf will be obtained between 5s<t<10 s.

Since, the back emf at t=3s is e,

the rate of change of current at t=3, s=slope of OA from t=0s to t=5s=1/5 A/s

So, Applying

ε = - L \frac{dI}{dt}

;

If u = Lx1/5

(for t = 3 s, \frac{dl}{dt} = 1 / 5)

(L is constant).

Step 2: Similarly, we have for other values:

For

5 s < t < 10 s u_{1} = - L \frac{3}{5} = - \frac{3}{5} L = - 3 e

Thus,

at t = 7 s, u_{1} = - 3 e

For 10s < t < 30s

u_{2} = L \frac{2}{20} = \frac{L}{10} = \frac{1}{2} e

For

t > 30 s, u_{2} = 0

Thus, the back emf at t = 7s, 15s and 40s are

- 3 e

, e/2 and 0 respectively.

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