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A 6.00 $\mu \mathrm{F}$ capacitor that is initially uncharged is connected in series with a 4500$\Omega$ resistor and a 500 $\mathrm{V}$ emf source with negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor, (b) the voltage drop across the resistor, (c) the charge on the capacitor, and (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities?

a. the voltage across the capacitor is zero just after the circuit completed.b. the voltage across the resistor is $500 \mathrm{V}$c. Initially, the charge on the capacitor is zero.d. the current passing through resistor is 0.111 $\mathrm{A}$e. the charge on the capacitor is 3.00 $\mathrm{mC}$

Physics 102 Electricity and Magnetism

Chapter 19

Current, Resistance, and Direct-Current Circuit

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Capacitance and Dielectrics

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Cornell University

University of Sheffield

University of Winnipeg

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in this case, we have secured that Looks like this. We have a sister 4,500 in parallel with her battery source on a capacitor. I'm sorry. In Siri's, the bathroom source under capacitor Pastor has the bastions of 600 micro car and the battery has an image for 500 Ward. Now we want to find out a few quantities as soon as he closed the switch. No, before we go ahead. Now that we're just close to switch, the charge on the capacitor is zero. That is obvious because the capacitors are charged and without any time passing, it can't accumulate any charge. Which means this cab faster at basically like a short. The current passes right through it, and it acts like just like a piece of wire, which means there is no child. There is no potential drop over it either. Andi current that passes through this circuit is simply affected by this resistor on B power source. So initially time zero, the circuit might as well be like this. Just a normal guy with her sister and the human. But what happens after a long period oftime once the capacity is completely charged. Once the capacitor is completely charged, the capacitor does not alone anymore. Current float to float through because the charge on it is constant Which means it basically acts like all open switch. So after a very long time, the circuit is equal in tow. This now that we have that let's proceed to the problem. Part is to find the oldest drop across the capacitor initially, because there is no charge on the capacitor. We know that voltage. We is Cuba Si So that is settled. I see just zero as we expected. There is no always drop across the capacitor. Or maybe we want to find out the oldest drop across the resistance because there is no whole days dropping plastic. A pastor all of the old age from the IMF must drop over there. Sister, which means the world is dropping costed her sister must be 4,500. What? This is even more clear. If you look at this picture, you can see that the gold is drop across. The resistor is equal to the world is drop across the battery for Patsy. We want to find out the current I'm sorry. The charge on the capacitor. This is simply a zero. The pastor has had no time to accumulate any charge and hence the charge would simply be zero and for party. We want to find out the current now current in this case because we know the circuit looks like this is simply we buy our that is 500 world. You had it before 1,005 100 home which is one by nine ampere which is 0.111 hand here. Now that we have this, let's do this for the second case after a very long time in a circle like this or by day, we want to find out he wanted to cross the capacitor. Now that there is this which is open, there is no current flowing through the circuit, which means all of the world age drop occurs across the switch because the whole day's job across the resistor. Zero. Because there is no current from Detroit between sparred B the whole day's drop across the resistor zero because that is equal to ay r on guys zero, which means all of the world is must now drop across the switch, which is the resistor hence the voltage across the pasture. I'm sorry. The capacitor The world is drop because the capacitor is all of it. The 500 would not for Potsie. I want to find out the charge. You know, the charge is we see in product ofthe board age times capacitance, which is 500 world multiplied by 600 into 10. Power minus six. This he's three times 10 bar minus three. Couldn't. Which turns out to be three micro. Colin, Um, before we go ahead, Yeah, The last part is the current, but we know. But the current in this problem must be zero, because this's just simply over on DH. There is all it is. Don't. It's simply zero. I'm sorry, but the give us a minute to 6.0 Micro fire. Thiss explained sirrah zero on bad his way. The number expected. Yes, for part B because the switches open, the capacitor is fully charged and it does not let any current in the current is simply zero

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