A circuit containing capacitors C1 and C2 as shown in figure are in steady state with key k1 closed at the instant t=0, if k1 is opened and K2 is closed then the maximum current in the circuit will be:
A
1 A
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B
5A
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C
2 A
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D
None of these
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Solution
The correct option is A1 A Fromthefigurevoltageacrosscapacitor2μF(c1)atsteadystate=10vwherek2isclosedMaxcurrentisobtainewhentotaleneegyofc1istransferedtotheinductorHence,12I2L=12cv2I=√cv22=√cLv=w√2×10−50.2×10−3=10√10×10−3=10−1‘×10=1A