1 in the given circuit where the switch closes at t0 s...

1. In the given circuit, where the switch closes at t=0 s, determine: a. The expression for the voltage in the capacitor b. The expression for the current in the capacitor c. The current and voltage of the capacitor at t=5 ms d. Find the numerical value of the energy stored in the capacitor as time approaches infinity, (remember the relation between power and energy) $R_1 = 240 \Omega$ $R_2 = 800 \Omega$ $R_3 = 200 \Omega$ $R_4 = 1040 \Omega$ $C_1 = 50 \mu F$ $V_1 = 100V$

Transcript

00:01 For this problem on the topic of direct current circuits, we are given a simple rc circuit and the values for the capacitance c, the resistance r and the emf epsilon are all given.

00:13 At the time t is equal to 10 seconds after the switch is closed, we want to calculate the charge in the capacitor, the current to the resistor, the rate at which the energy is being stored in the capacitor, and the rate at which energy is being delivered by the battery.

00:28 So in part a to calculate the charge q on the capacitor at this instant, we know this is equal to c delta v into 1 minus e to the minus t over rc.

00:46 And these values are all known so we can put them in as follows.

00:50 That's the capacitance of 1 times 10 to the minus 6 ferrets times delta v, which is 10.

01:00 Volts multiplied by 1 minus e to the minus 10 seconds divided by the resistance, which is 2 times 10 to the power 6 oms times the capacitance, 1 times 10 to the minus 6 ferrets.

01:31 So if we calculate this, we get the charge in the capacitor after 10 seconds to be 9 .93 microcolums.

01:49 And that's our solution for part a.

01:53 For part b, we want to find the current through the resistor.

02:00 So we know the current through the resistor i is defined as dqdt, which we can write as delta v.

02:13 Over r multiplied by e to the minus t over r c and again all these values are known so if we substitute them in we get this to be 10 volts divided by 2 times 10 to the power 6 oms which is the resistance times e to the power minus 5 and so calculating this becomes 3.

02:51 0 .37 times 10 to the minus 8 ampiers, which you can write as 33 .7 nanopheres...

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