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willydavidjr
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Consider the electrical circuit as shown on the website I provided below(and an attachment I provided too). Consisting of E=6[V] battery, two switches S1 and S2, two resistors R1=4ohms and R2=2ohms, and a capacitor C=2 microFarad. The internal resistance of the battery maybe ignored. Initially the switches are both open and the capacitor has no charge. Close the S1 at a certain time. At a sufficiently long time after the S1 is closed, the capacitor is fully charged and the circuit becomes steady.
Question:
1.) Just after the S1 is closed, what is the current flowing through R1.
2.) How much charge is stored in the capacitor C?
3.) During the period in which the capacitor is charged, how much work done by the battery.
My answer:
1.)1.5 amperes
2.)From q=CV
so 2*6= 12 micro Coulombs
3.)From E=1/2CV^2
so 1/2(2)(6^2)
=36 micro joules.
Am I correct?
This is the website: www.geocities.com/willydavidjr/circuit.html
Question:
1.) Just after the S1 is closed, what is the current flowing through R1.
2.) How much charge is stored in the capacitor C?
3.) During the period in which the capacitor is charged, how much work done by the battery.
My answer:
1.)1.5 amperes
2.)From q=CV
so 2*6= 12 micro Coulombs
3.)From E=1/2CV^2
so 1/2(2)(6^2)
=36 micro joules.
Am I correct?
This is the website: www.geocities.com/willydavidjr/circuit.html