Asked by gOOD eVE
A 2.0-microF capacitor is charged to 50V and then connected in parallel (positive to positive plate) with a 4.0-microF capacitor charged to 100V. (a) What are the final charges on the capacitors? (b) What is the difference across each?
Answers
Answered by
bobpursley
voltages will be equal.
q=C V charge on first: 2E-6&50
charge on second: 4E-6*100 so the charges total is (added)5E-6*100
so in the final condiguration, qt=sum of charges above=5E-6*100
and the voltages are equal.
total capacitance in parallel is 6E-6, so
V=q/C=5E-6*100/6E-6 =500/6 volts
then the final charges on each capacitor is
q1=C1*v
q2=C2*v
q=C V charge on first: 2E-6&50
charge on second: 4E-6*100 so the charges total is (added)5E-6*100
so in the final condiguration, qt=sum of charges above=5E-6*100
and the voltages are equal.
total capacitance in parallel is 6E-6, so
V=q/C=5E-6*100/6E-6 =500/6 volts
then the final charges on each capacitor is
q1=C1*v
q2=C2*v
Answered by
vicky
q=cu
find the charge of the first capacitor
the find the equivalent capacitance of the parallel and q=c<?>(100)
U(AB)=VA-VB
find the charge of the first capacitor
the find the equivalent capacitance of the parallel and q=c<?>(100)
U(AB)=VA-VB
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