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Answer :
Explanation:
The give data is as follows.
C = 6 [tex]\mu F[/tex] = [tex]6 \times 10^{-6} F[/tex]
V = 6 V
Now, we know that the relation between charge, voltage and capacitor for series combination is as follows.
Q = CV
= [tex]6 \times 10^{-6} F \times 6 V[/tex]
= [tex]36 \times 10^{-6} C[/tex]
or, = 36 [tex]\mu C[/tex]
Thus, we can conclude that maximum charge of the given capacitor is 36 [tex]\mu C[/tex].
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Answer:
Explanation:
capacitance, C = 6 μF
Voltage, V = 6 V
Let the maximum charge is Q.
Q = C x V
Q = 6 x 6 = 36 μC
