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A 12.4 $\mu \mathrm{F}$ capacitor is connected through a 0.895 $\mathrm{M\Omega}$ resistor to a constant potential difference of 60.0 $\mathrm{V}$ . (a) Compute the charge on the capacitor at the following times after the connections are made: $0,5.0 \mathrm{s}, 10.0 \mathrm{s}, 20.0 \mathrm{s},$ and 100.0 $\mathrm{s}$ . (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for $t$ between 0 and 20 $\mathrm{s}$ .

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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

University of Michigan - Ann Arbor

University of Winnipeg

McMaster University

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In the arc circuit, capstone of the capisti is 12.4 microfiso. Here we have capstan equal to 12 fit 4 micropharet, which is equal to 12.4 times 10 to the power minus 6 ver resistance of the resister is equal to 0.895 mega, which is equal to 0.895 times 10 to the power 6 on and impis 60 vol pot. Now, in the first part of this problem, we want to calculate the charge stored on the capital. At t is equal to 0. At t is equal to 5 seconds at t is equal to 10 seconds at t is equal to twenty second and a equal to one hundred. Second, it as we have the relation for ja stored on a capester is come, is equal to copestone times g into 1 minus e to the power minus time. Upon time, constant now at t is equal to 0. We have just stored on captare is equal to c times v into 1 minus e to the power zent, so we have c equal to 0. Now t is equal to 5 seconds. We have at 5 seconds equal to her. Capacitance is 12.4 times 10 to the power minus 6 ferret mpabattery is 60 volt into 1 minus a near times minus 5 seconds upon r c. Here, r is 0 point. A 95 multiplied 10 to the power 6 or multiply c is 12 multiply, 10 to the power minus 6 part. So we have charge at 5 seconds equal to 12 time 4 times 10 to the power minus 6 into 6 part 0 volt into 1 minus 0.6372. Eightnt, so by multiplying all these values we get just tore on at 5 seconds on the cap, star is equal to 2.7 times 10 to the power minus 4 coulomb. Now it t is equal to 10. Second, tension is equal to here resistance here, capacitance of the capitol is 12.4 times 10 to the power. Minus 6 fat imports, 60 vol into 1 minus minus 2 upon time constant here time, constant is the product of r and c, which is equal to 11.098. Second, so, while computing, all these numbers we get just stored in a capstern t is equal to 10. Seconds is 4.42 times 10 to the power minus 4. Coulomb now t is equal to twenty. Second, we have q. Twenty second is equal to c times here. Capstones 12.4 times 10 to the power. Minus 6 were mefaits 6 into 1 minus e to the power minus time here times, twenty second upon time, constant, which is 11.098 seconds, so we have c. Twenty second is equal to 6.21 times 10 to the power minus 4. Colomnow at t is equal to 100 seconeequal to 12.4 times 10 to the power minus 6 pi into m f, a d 6.0 vol into 1 minus e to the power time here. Time is one hundred second upon time. Constant is 11.098 coso. We have charge at one hundred. Second is equal to 7.404 times 10 to the power minus 4 coulomb. Now we are going to solve b part of this problem in b. Part of this problem we want to calculate the current at t is equal to 0. At t is equal to 5 is equal to 20 and is equal to one hundred. Second, as we have the relation for current in our circuit, is it is equal to m fog upon resistance into e to the power minus time upon time. Constant now t is equal to 0. Second, we have i at 0 equal to her is 60.0 volt upon resistance is 0.95 times 10 to the power 6 here e to the power minus t upon time here time. So we have e to the power 0 point. So we get corn at t is equal to 0 is 6.7 times 10 to the power minus 5. Ampere now t is equal to 5 seconds. We have rent at 5 seconds is equal to v. Upon r e v is 60 volt upon resistance is 0.895 times e to the power minus t upon time. Constant here time is minus 5 seconds upon time. Constant, the product of r c is 11.098 seconds, so we have got at 5 seconds equal to 4.27 times 10 to the power minus 5 ampere now tis equal to 10 seam, we have a tension, is equal to m f of t 60 volt upon resistance is 0 point n, 95 times 10 to the power 6 into e to the power minus time 10 seconds upon time, constant, which is 11.098 seconds, so we have gotten second equal to 2.72 times 10 to the power minus 5 ampere now t is equal to twenty. Second, we have an equal to 6 volt upon 0.895 times 10 to the power 6 on into e to the power minus twenty second upon time, constant is 11.098 second, so we have current at twenty second equal to 1.11 times 10 to the power minus 5 ampere. Now a t is equal to under secondhand 800 seconds is equal to 60 volt upon 0.895 times 10 to the power 6 into e to the power minus one hundred second upon 11.098. Second, so we have current eight hundred second equal to 8.18 times 10 to the power minus 9 ampere. Now in c part of this problem, we want to sketch the graphs between time and charge and graph between time and current for charging of capesterhere. We have taken time along x, axis and charge along y axis at t is equal to 0. Just stored on incapacity is 0. Now at t is equal to 5. Second, we have charge of the capstor equal to 2.7 times 10 to the power minus 4 coulombe. So, at 5 seconds, weaveequal to 2.7 times 10 to the power minus 7 at isa just stored in the capacities 4.42 times 10 to the power minus 4 colomba is equal to twenty second point: we have just stored in the cepisti 6.21 times 10 to the power Minus 4 coulomb, so tis equal to hundreds, have charge on the capsture equal to 7.404 times 10 to the power minus 4 coulombs. When we join all these points, then we get a curve, so here this curve shows that this store on the store increases exponentially now. Here we have taken rental time along x axis, and the current along y axis at the beginning at t is equal to 0 point. We have maximum current in the circuit that is 6.7 times 10 to the power minus 5 ampere after 5 seconds. The current reduces to 4.2 times 10 to the power minus 5 ampere. So after 10 seconds we have corn equal to 2.72 times 10 to the power minus 5 ampere at 27, equal to 1.11 times 10 to the power minus 5 ampere. So after one hundred second, we have current equal to 0.000818 times 10 to the power minus 5 ampere. So after one hundred second, we have almost practical current equal to 0. So when we join all these points on the graph, they will get a exponentially decreasing graph.

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