A capacitor that is charged to a potential of 12.0 V is then connected to a voltmeter that has an internal resistance of&nbsp;3.40MΩ. After a time of 4.00 s the voltmeter reads 3.0 V.&nbsp;What are (a) the capacitance and (b) the time constant of the circuit?

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&amp;nbsp;3.40MΩ. After a time of 4.00 s the voltmeter reads 3.0 V.&amp;nbsp;What are (a) the capacitance and (b) the time constant of the circuit?

esfloravaou · Accepted Answer

Step 1  Knowns  R−C circuit, the charge on a discharging capacitor as a function of times is given by:  1) q=Qfe−tRC  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  Given:  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  Calculations  Solving equation (2) for V, we get:  V=QC  So, by dividing equation (1) by C, we get the voltage across a discharging capacitor as a function of ti,e:  3) V=V0e−tRC  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 · Accepted Answer

Step 1&amp;nbsp;This is the discharge of electricity through a resistor, so the decrease in voltage and charge is exponentialV=V0e−tRC&amp;nbsp;ln⁡VV0=−tRC&amp;nbsp;RC=−tln⁡(VV0)=tln⁡(V0V)&amp;nbsp;RC=4ln⁡(123)=2.885&amp;nbsp;seconds&amp;nbsp;Thats

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A capacitor is charged to a potential of 12.0 V

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