A) 441 pF
B) 242 pF
C) 242 nF
D) 441nF
Correct Answer: B
Solution :
Option [b] is correct. Explanation: The value of the capacitor marked as 221 is 220 pF. The value of the capacitor marked as 220 is 22 pF. When connected in parallel, the total capacitance = 220 pF + 22 pF =242pF.You need to login to perform this action.
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