jubateee

2022-01-03

A capacitor that is charged to a potential of 12.0 V is then connected to a voltmeter that has an internal resistance of $3.40M\mathrm{\Omega }$. After a time of 4.00 s the voltmeter reads 3.0 V.
What are (a) the capacitance and (b) the time constant of the circuit?

esfloravaou

Step 1
Knowns
$R-C$ circuit, the charge on a discharging capacitor as a function of times is given by:
1) $q={Q}_{f}{e}^{-\frac{t}{RC}}$
Where ${Q}_{f}$ is the initial charge on the capasitor and $\tau =RC$ is the time constant of the circuit.
The capacitance C of a capacitor is the ratio of the chatge to the potential difference betwen its plates:
2) $C=\frac{Q}{V}$
Step 2
Given:
The initial voltage difference across the capacitor is ${V}_{0}=12.0V$, the internal resistance of the valtmater is $R=3.40×{10}^{6}\mathrm{\Omega }$ and after time $t=4.00s$
Step 3
Calculations
Solving equation (2) for V, we get:
$V=\frac{Q}{C}$
So, by dividing equation (1) by C, we get the voltage across a discharging capacitor as a function of ti,e:
3) $V={V}_{0}{e}^{-\frac{t}{RC}}$
Where ${V}_{0}=\frac{{Q}_{f}}{C}$ is the initial voltage diffence.
When we connent the capacitor to the voltameter, the capacitor discharges its charge in the internal resistance of the voltameter.
And from Kirchoffs

Orlando Paz

Step 1
This is the discharge of electricity through a resistor, so the decrease in voltage and charge is exponential
$V={V}_{0}{e}^{-\frac{t}{RC}}$
$\mathrm{ln}\frac{V}{{V}_{0}}=-\frac{t}{RC}$
$RC=-\frac{t}{\mathrm{ln}\left(\frac{V}{{V}_{0}}\right)}=\frac{t}{\mathrm{ln}\left(\frac{{V}_{0}}{V}\right)}$
$RC=\frac{4}{\mathrm{ln}\left(\frac{12}{3}\right)}=2.885$ seconds
Thats

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