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The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
<h3>
Total capacitance of the circuit</h3>
The total capacitance of the circuit is calculated as follows;
Capacitors in series;
1/Ct = 1/8 + 1/7.5
1/Ct = 0.25833
Ct = 3.87 mF
Capacitors is parallel;
Ct = 3.87 mF + 12 mF + 15 mF
Ct = 30.87 mF
Ct = 0.03087 F
<h3>Charge in each capacitor</h3>
Q = CV
Q = 0.03087 x 12
Q = 0.37 C
Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
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Answer:
A
Explanation:
R=V/I
IF you double the resistance, it's become:
2R=(1/2)I
Answer:
B type ii supernova is the right answer