Question

1) In the next circuit, the battery has a potential difference of 20 V and it is connected to the four capacitors as shown. Find the charge and voltage on each capacitor (0.5p) 2 uF 4 F 5 F 12 F 2) In an RC charging circuit, Vs = 200 V, R = 2 x 10 Ohm, C = 50 uF and Qe = 0 at t = 0. Find: (1.5p) (a) the potential difference across C after one second; (b) the current in R after 2 seconds; (c) the time needed for the Voltage on the capacitor to be equal to the Voltage on the resistor

Answer 1

T y 1 三十 工会 - + 20v Fig 2 5 一等 一章--- tia 4 *
1. The circuit consists of parallel and series capacitors. Capacitance is the charge stored divided by the voltage passing through it. This is give mathematically as,

$C=\frac{Q}{V}$

In a series capacitors, the charge and the voltage flowing through the capacitors is the same.

Hence the final Capacitance given by C is

$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+.....+\frac{1}{C_{n}}$

where 1,2,3...n refer to the number of capacitors.

If capacitors are placed parallel the resulting capacitance is given by

C = Ci+ C2 + C3 + ....... + Cn

This is because the voltage across each of them is the same .

Now we can simply the circuit with this knowledge. The $2\mu F$ and $4\mu F$ are in series. So using the formula we get the resulting Capacitance for these two as

$\frac{1}{C}=\frac{1}{2}+\frac{1}{4}$

$C=\frac{4}{3}\mu F$

This resultant capacitance is now parallel to $12 \mu F$ . So we can simply add

1. The circuit consists of parallel and series capacitors. Capacitance is the charge stored divided by the voltage passing through it. This is give mathematically as,

C=(Q/V)

In a series capacitors, the charge and the voltage flowing through the capacitors is the same.

Hence the final Capacitance given by C is

$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+.....+\frac{1}{C_{n}}$

where 1,2,3...n refer to the number of capacitors.

If capacitors are placed parallel the resulting capacitance is given by

C = Ci+ C2 + C3 + ....... + Cn

This is because the voltage across each of them is the same .Now we can simply the circuit with this knowledge. The 2micro F and 4 micro F are in series. So using the formula we get the resulting Capacitance for these two as
$\frac{1}{C}=\frac{1}{2}+\frac{1}{4}$

$C=\frac{4}{3}\mu F$

This resultant capacitance is now parallel to 12 micro F. So we can simply add them to get 12 micro F +(4/3) micro F =(40/3) micro Farad
This resultant capacitance is in series with 5 \mu F so we use the series formula to get

$\frac{1}{C}=\frac{1}{5}+\frac{3}{40}$

them to get $12 \mu F +\frac{4}{3} \mu F=\frac{40}{3} \mu F$

This resultant capacitance is in series with $5 \mu F$ so we use the series formula to get

$\frac{1}{C}=\frac{1}{5}+\frac{3}{40}$

$C=\frac{40}{11}\mu F$

So this is the equivalent final Capacitance of the Circuit.

The voltage is given to be 20 V.

Hence total charge is Q=CV

$Q=\frac{800}{11}\mu F$

Refer image for figures.

We use figure 3 to find charge and voltage across the $5\mu F$

The charge is the same as total charge so by C=(Q/V)

V for $5\mu F$ =Q/C​​​​​​1

=800/11x5

=160/11 V =14.55 V

V for the 40/3 micro Farad=Q/C

=(800x3)/(11x40)

=220/3 micro Farad

Now we use Fig 2 to find remaining answers. Since the 12 micor Farad is parallel to (4/3) micro Farad, the voltage across them is the same but the charge for each of them is different. The capacitor with higher Capacitance gets more charge

V=220/3

So charge on (4/3) micro Farad =(4x220)/(3x3)

=(880/9)

Charge on 12 micro Farad=(12x220)/3

=880

Now we use figure 1,

The charge for the 2 and 4 micro Capacitance is the same. Voltage for each is calculated.

Q=880/9

V for 2 micro Farad=(880/9x2)

=(440/9) V

=48.89 V

V for 4 micro Farad=(880/9x4)

=(220/9)

=24.44 V

Hence we found charge and voltage for each.

2. Voltage across a charging capacitor is given by

$V=V_{input}(1-e^{-t/RC})$

Since charging takes place exponentially and is dependent on R and C

V at input is 200 V so we substitute the values in the question to get

V at 1s for capacitor=200(1-e(-1/10))

=19V

B. Charging current across a capacitor is given by

$I=\frac{V_{input}}{R}e^{-t/RC}$

So upon substitution we get,

I=10-3.e​​​​​-2/10

=8.187x10-4 A

c. Voltage across resistor is given by

$V={V_{input}}e^{-t/RC}$

Voltage across capacitor is give by

$V={V_{input}}(1-e^{-t/RC})$

We need to find the time when these two are equal,

So ${V_{input}}e^{-t/RC}={V_{input}}(1-e^{-t/RC})$

$e^{-t/RC}=\frac{1}{2}$

Putting in the values of R and C we get

t=6.9s

Hence this is the time when the voltage across capacitor is equal to voltage across resistor.