The current i in the circuit (see figure) is :

\frac{1}{45} A

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\frac{1}{15} A

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\frac{1}{10} A

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\frac{1}{5} A

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Solution

The correct option is C $\frac{1}{10} A$
R.E.F.Image.
Apply KVL Law from ODAB
$2 - 30 i_{1} = 0$
$i_{1} = 2 / 30 = 1 / 15 A M P$
Now apply KVL in ODA CBO
$2 - 30 (i - i_{1}) - 30 (i - i_{1}) = 0$
$2 = 60 (i - i_{1})$
$\frac{1}{30} = (i - i_{1})$
$i = \frac{1}{15} = \frac{1}{30}$
$i = \frac{1}{30} + \frac{1}{15} = \frac{3}{30}$
$i = \frac{1}{10} A M P$

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A 10 henry inductor is carrying a current 2A. Rate at which the current be changed so that an e.m.f. of 100V is induced is

V_{B E} = 0.7 V

5 \times 10^{- 15} A

V_{T} = 26 m V, η = 1

i_{1} / i_{2}

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