In the circuit shown in the following figure E = 10 V, \[{{R}_{1}}=2\,ohm,\] \[{{R}_{2}}=3\,ohm\] and \[{{R}_{3}}=\text{6}\,ohm\]and L = 5 henry. The current \[{{I}_{1}}\] just after pressing the switch S is |
A) 2.5 amp
B) 2 amp
C) 5/6 amp
D) 5/3 amp
Correct Answer: B
Solution :
just after closing of switch inductor behave as open circuit equivalent circuit is shown |
\[{{I}_{1}}=\frac{E}{{{R}_{1}}+{{R}_{2}}}=\frac{10}{2+3}=\frac{10}{5}=2mp\] |
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