Capacitor Mathematics

Question 1

what is the equation for the time constant

Answer 1

T = CR

Question 2

what do each of those variables stand for

Answer 2

• T = time constant
• C = capacitance
• R = resistance (in circuit)

Question 3

what is the equation for the remaining charge on a capacitor that is discharging through a resistance of R

Answer 3

Q(t) = Q(0) * e^(-t / CR)

Question 4

what do each of those variables stand for

Answer 4

• Q(t) = charge at time t
• Q(0) = initial charge (at t = 0)
• t = time
• CR = time constant T
• e = the constant (2.718…)

Question 5

a 0.03F capacitor is fully charged by a 12V supply and is then connected to discharge through a 900 ohm resistor. how would you work out how much charge remains on the capacitor after 20 seconds

Answer 5

• youre using Q(t) = Q(0) e^(-t / CR)
• but you need Q(0) first
• C = Q / V so Q(0) = CV which is 0.03 x 12
• you also need CR (T) which is just 0.03 x 900
• with Q(0), -t and CR you can work out Q(20)

Question 6

what is the equation for the remaining pd across a capacitor that is discharging through a resistance of R

Answer 6

V(t) = V(0) * e^(-t / CR)

Question 7

what is the equation for the remaining current in a capacitor that is discharging through a resistance of R

Answer 7

I(t) = I(0) * e^(-t / CR)

Question 8

what would an increase in time do the values of remaining Q, V and I

Answer 8

an increase in time would result in the exponential decrease of all of them

Question 9

why does an increase in time lead to the exponential decrease in Q, V and I when looking at the equation for each one

Answer 9

• an increase in time leads to an ‘increase’ in the -ve value of t in ^(-t / CR)
• this results in an ‘increase’ in the magnitude of that -ve fraction (technically decrease due to -ve sign)
• which makes e^(-t / CR) aka 1 / e^(t / CR) smaller
• multiplying it with the initial Q V or I would therefore lead to a decrease in V, Q or I(t) as time goes on

Question 10

how would you mathematically get the equation for remaining pd from the equation for remaining charge (this is long af just know what method is needed)

Answer 10

• by taking logs of both sides and replacing Q with CV
• using Q = Q(0) * e^(-t / CR)
• lnQ = ln(Q(0) * e^(-t / CR))
• lnQ = lnQ(0) + ln(e^(-t / CR))
• lnQ = lnQ(0) - t / CR
• Q = CV, so
• ln(CV) = ln(CV) -t / CR
• lnC + lnV = lnC + lnV(0) - t / CR
• capacitance is constant, so subtract lnC from both sides
• lnV = lnV(0) - t/CR
• e^lnV = e^(lnV(0) - t/CR)
• V = e^lnV(0) * e^(-t/CR)
• V = V(0) * e^(-t / CR)

Question 11

what is the ‘half-life’ of a capacitor when discharging

Answer 11

the time it takes for the charge on a capacitor to reach 37% of its initial value

Question 12

why is the number 37% and not 50%

Answer 12

• because in the maths for working out Q when time = CR (its ‘half-life’ point)
• e^(-t / CR) would = e^(-CR / CR)
• = e^-1 = 1 / e
• and 1 / e approximates to 37%

Question 13

what needs to be noted about how the charge , pd and current across a capacitor overtime compare for charging and discharging

Answer 13

• with charge and pd, charging would result in a ‘decreasingly’ exponential increase and discharging would result in a ‘decreasingly’ exponential decrease
• btw ‘decreasingly’ just means the lines level out
• whereas for current, it exponentially decreases in both charging an discharging scenarios
• so its basically the odd one out

Question 14

what would be the equations for remaining Q and V when charging

Answer 14

• V(t) = V(0) * (1 - e^(-t/CR))

- Q(t) = Q(0) * (1 - e^(-t/CR))