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A capacitor of $5\ \mu$F is connected across a $12$ V battery. The charge stored is:
A$5\ \mu$C, equal to the capacitance value alone
B$2.4\ \mu$C, dividing voltage by capacitance
C$60\ \mu$C, since $Q = CV = 5\cdot 12 = 60$
D$17\ \mu$C, the simple sum of the two values
Answer & Solution
Correct answer: C. $60\ \mu$C, since $Q = CV = 5\cdot 12 = 60$
$Q = CV = 5\ \mu\text{F}\cdot 12\text{ V} = 60\ \mu$C.
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