The circuit in the figure below contains a 9.00 V battery and four capacitors. The two capacitors on the left and right both have same capacitance of C, = 8.40 µF. The capacitors in the top two branches have capacitances of 6.00 µF and C, = 2.40 µF. 6.00 μF + 9.00 V (a) What is the equivalent capacitance (in µF) of all the capacitors in the entire circuit? |HF (b) What is the charge (in uC) stored by each capacitor? right 8.40 µF capacitor left 8.40 µF capacitor 2.40 µF capacitor 6.00 µF capacitor (c) What is the potential difference (in V) across each capacitor? (Enter the magnitudes.) right 8.40 µF capacitor V left 8.40 µF capacitor 2.40 µF capacitor V 6.00 µF capacitor

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The circuit in the figure below contains a 9.00 V battery and four capacitors. The two capacitors on the left and right both have same
capacitance of C, = 8.40 µF. The capacitors in the top two branches have capacitances of 6.00 µF and C, = 2.40 µF.
6.00 µF
C
9.00 V
(a) What is the equivalent capacitance (in uF) of all the capacitors in the entire circuit?
(b) What is the charge (in µC) stored by each capacitor?
right 8.40 µF capacitor
left 8.40 µF capacitor
2.40 µF capacitor
6.00 µF capacitor
(c) What is the potential difference (in V) across each capacitor? (Enter the magnitudes.)
right 8.40 µF capacitor
V
left 8.40 µF capacitor
2.40 µF capacitor
V
6.00 µF capacitor
V
Transcribed Image Text:The circuit in the figure below contains a 9.00 V battery and four capacitors. The two capacitors on the left and right both have same capacitance of C, = 8.40 µF. The capacitors in the top two branches have capacitances of 6.00 µF and C, = 2.40 µF. 6.00 µF C 9.00 V (a) What is the equivalent capacitance (in uF) of all the capacitors in the entire circuit? (b) What is the charge (in µC) stored by each capacitor? right 8.40 µF capacitor left 8.40 µF capacitor 2.40 µF capacitor 6.00 µF capacitor (c) What is the potential difference (in V) across each capacitor? (Enter the magnitudes.) right 8.40 µF capacitor V left 8.40 µF capacitor 2.40 µF capacitor V 6.00 µF capacitor V
Expert Solution
Step 1

(a)

C2 and 6.00 μF are parallel. Therefore,

Cp=C2+6.00 μF=2.40 μF+6.00 μF=8.40 μF

Two C1 and Cp are in series. Therefore, equivalent capacitance is,

1C=1C1+1C1+1Cp1C=18.40 μF+18.40 μF+18.40 μFC=2.80 μF

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