Problem 19.46 A 0.50-μF and a 1.4-μF capacitor (C1 and C2, respectively) are connected in series to a 14-Vbattery. Part A Calculate the potential difference across each capacitor. (in volts) Part B Calculate the charge on each capacitor. (in Coulombs) Part C Calculate the potential difference across each capacitor assuming the two capacitors are in parallel. (in Volts) Part D Calculate the charge on each capacitor assuming the two capacitors are in parallel. (in Coulombs) Thanks for the help! If you end up handwriting your answer, please do so in super clear handwriting! |
Here ,
for the capacitor in series
1/Ceq = 1/C1 + 1/C2
1/Ceq = 1/0.50 + 1/1.4
Ceq= 0.368 uF
part A)
potential difference across C1 = C2 * V/(C1 + C2)
potential difference across C1 = 1.4 * 14/(0.50 + 1.4)
potential difference across C1 = 10.3 V
potential difference across C2 = 14 - 10.3 V
potential difference across C2 = 3.68 V
part B)
as the charge on each capacitor is same in series
Qeq = Ceq * V
Qeq = 14 * 0.368 uC
Qeq = 5.15 uC
the equivalent capacitance is 5.15 uC
part C)
as the potential is same in both capacitors
potential across capacitor C1 and C2 = 14 V
part D)
for C1 ,
charge on C1 = C1 * V
charge on C1 = 0.50 * 14 uC
charge on C1 = 7 uC
for C2
charge on C2 = C2 * V
charge on C2 = 1.4 * 14
charge on C2 = 19.6 uC
Problem 19.46 A 0.50-μF and a 1.4-μF capacitor (C1 and C2, respectively) are connected in series...
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Two capacitors, C1 = 4.92 μF and
C2 = 14.1 μF, are connected in
parallel, and the resulting combination is connected to a 9.00-V
battery.
(a) Find the equivalent capacitance of the combination.
(b) Find the potential difference across each capacitor.
(c) Find the charge stored on each capacitor.
*PLEASE ANSWER ALL PARTS TO A, B, AND C CLEARLY* THANK YOU FOR
YOUR HELP IN ADVANCE!
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