Q.

In the circuit shown, charge on the 5μF  capacitor is :

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a

16.36 μC

b

5.45 μC

c

10.90 μC

d

18.00 μC

answer is A.

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Detailed Solution

Applying KJR  q=q1+q2Applying KLR  6−q12−q5=0And 6−q24−q5=0⇒q2=2q1​⇒q=3q1​Thus , ​6−q6−q5=0​⇒1130q=6​⇒q=18011μC=16.36μC
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