5) You connect a battery, resistor, and capacitor as in Fig....

5) You connect a battery, resistor, and capacitor as in Fig. a, where ? = 36.0V, C = 5.00?F, and R = 120?. The switch, S, is closed at t = 0 in Fig b. a) (5 pts) When the voltage across the capacitor is 8.00 V, what is the magnitude of the current in the circuit? b) (5 pts) When the voltage across the capacitor is 8.00 V, at what rate is energy being stored in the capacitor? c) (10 pts) At what time t after the switch is closed is the voltage across the capacitor 8.00 V? (a) Capacitor initially uncharged Switch open ? i = 0 q = 0 R C (b) Charging the capacitor Switch closed ? i +q -q R C

Transcript

00:01 All right, dear learners, let's draw the circuit diagram first.

00:08 Here we have an emf source.

00:13 With this emf source, we have a switch here.

00:17 With this switch, we have a capacitor connected and a resistor is connected in series.

00:25 This is finally connected to the emf source.

00:29 Let's label this diagram.

00:31 This emf source, okay, it is given that it is a second.

00:35 This emf source it is of 36 volts this capacitor the capacitance of this capacitor it is given it is of 5 micro ferrard all right and the resistance of this resistor is given it is of 120 oms and this switch is let's say s all right to solve part a we will use kvl or kirkov's voltage law in this loop.

01:16 According to this law, some of all the potentials across in the close path or in the loop should be equal to 0.

01:26 All right, first of all, we'll start from here this potential first, then the potential across this resistor, let's say the current is i minus i times this resistance, this potential, then the potential drop across this capacitor, let's denote that with vc is equal to zero now plugging the values this emf source it is given that it is of 36 volts minus i the value of resistance is given that is 120 oms minus the value of capacitor the voltage of the capacitor is given this is 8 volts it is equal to 0 from here i is equal to 28 volts divided by 120 oms so it means that the current flowing through the circuit when the capacitor acquires a voltage of 8 volt comes out to be 0 .0 to 3 .3 ampers now let's move to part b all right rate of energy flowing through the capacitor when the capacitor acquires a voltage of 8 volts.

02:55 For the 8 we will use the simple formula...

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