A capacitor is charged to a potential of 12.0 V



Answered question


A capacitor that is charged to a potential of 12.0 V is then connected to a voltmeter that has an internal resistance of 3.40MΩ. 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?

Answer & Explanation



Beginner2022-01-04Added 43 answers

Step 1
RC circuit, the charge on a discharging capacitor as a function of times is given by:
1) q=QfetRC
Where Qf is the initial charge on the capasitor and τ=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=QV
Step 2
The initial voltage difference across the capacitor is V0=12.0V, the internal resistance of the valtmater is R=3.40×106Ω and after time t=4.00s
Step 3
Solving equation (2) for V, we get:
So, by dividing equation (1) by C, we get the voltage across a discharging capacitor as a function of ti,e:
3) V=V0etRC
Where V0=QfC 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

Orlando Paz

Beginner2022-01-05Added 42 answers

Step 1 
This is the discharge of electricity through a resistor, so the decrease in voltage and charge is exponential
RC=4ln(123)=2.885 seconds 

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