Given a 1.75 uF capacitor, a 7.50 uF capacitor, and a 7.50 V battery, find the charge on each capacitor if you connect them in the following ways. (0) in series across the battery 1.75 uF capacitor 7.50 uF capacitor Juc (b) in parallel across the battery 1.75 uF capacitor 7.50 uF capacitor

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Given a 1.75 uF capacitor, a 7.50 uF capacitor, and a 7.50 V battery, find the charge on each capacitor if you connect them in the following ways.
(a) in series across the battery
Juc
Juc
1.75 uF capacitor
7.50 uF capacitor
(b) in parallel across the battery
1.75 uF capacitor
7.50 uF capacitor
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Transcribed Image Text:Given a 1.75 uF capacitor, a 7.50 uF capacitor, and a 7.50 V battery, find the charge on each capacitor if you connect them in the following ways. (a) in series across the battery Juc Juc 1.75 uF capacitor 7.50 uF capacitor (b) in parallel across the battery 1.75 uF capacitor 7.50 uF capacitor Need Help? Read It
Expert Solution
Step 1 For the capacitors are connected in series across the battery

When the capacitors are connected in series, the charge stored in each capacitors are same but the potential drop across each capacitor is different. 

The equivalent capacitance is given by 

1Cs=1C1+1C2    Here CS= series equivalent capacitance, C1=1.75 μF,  C2=7.5μF

1Cs=C1+C2C1C2Cs=C1C2C1+C2=1.75μF×7.5μF1.75μF+7.5μF=13.1259.25μF=1.419μF

Now let Q be the charge stored in each capacitor, then

Q=CsV     here V is the potential drop of the battery and V=7.5 V

Q=1.419μF×7.5 V=10.64μC

Since the charge stored in each capacitor are same, 

Charge stored across C1 and C2 are same and is equal to 10.64μC

 

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