In the circuit shown, the switch S is open for a long time and is closed at t = 0. The current i(t) for t≥0+ is
A
i(t)=0.5−0.125e−1000tA
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B
i(t)=1.5−0.125e−1000tA
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C
i(t)=0.5−0.5e−1000tA
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D
i(t)=0.375e−1000tA
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Solution
The correct option is Ai(t)=0.5−0.125e−1000tA Method 1:
At t=0−, the circuit is shown below:
Current will divide equally in both 10Ω resistors as shown in diagram.
At t≥0+, the circuit in s-domain is shown below: [L=15mH,i(0+)=i(0−)=0.75A]
The above circuit can be rearranged as given below:
Applying nodal analysis, V(s)−1.5s10+V(s)10+V(s)+15×10−3×0.7510+15×10−3s=0 V(s)10+V(s)10+V(s)10+15×10−3s =1510s−15×10−3×0.7510+15×10−3s V(s)[10+15×10−3s+55(10+15×10−3s)] =15[10+15×10−3s−10s×10−3×0.7510s(10+15×10−3s)] (1+10−3s)V(s)=(10+7.5×10−3s)2s[s+10001000] V(s)=10000+7.5s2000s V(s)=10000+7.5s2s(s+1000) =100002×1000s+{10000+7.5(−1000)}2(−1000)(s+1000) =5s−1.25s+1000
Taking inverse Laplace transform, v(t)=5−1.25e−1000tV i(t)=V(t)10 =0.5−0.125e−1000tA
Method 2: i(0+)=0.752=0.375A i(∞)=0.5A i(t)=i(∞)−{i(∞)−i(0+)}e−Rt/L
where, R = equivalent resistance seen across L with current source opened R=10+(10||10)=15Ω i(t)=0.5−{0.5−0.375}e−15t/15×10−3 =0.5−0.125e−1000tA