A capacitor is charged to a potential of 12.00 V and is then connected to a voltmeter having an internal resistance of 3.40 MΩ. After a time of 4.00 s, the voltmeter reads 3.00 V. What are (a) the capacitance and (b) the time constant of the circuit?

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user_27qwe · Accepted Answer

To solve this problem, we can use the formula for the voltage across a charging capacitor:V(t)=V0(1−e−tRC)where:- V(t) is the voltage across the capacitor at time t- V0 is the initial voltage across the capacitor- R is the resistance in the circuit- C is the capacitance of the capacitor- e is the base of the natural logarithm(a) To find the capacitance (C) of the circuit, we can rearrange the formula:C=−tR·ln(V(t)V0−1)Given:- V0=12.00 V (initial potential)- V(t)=3.00 V (voltage after 4.00 s)- R = 3.40 MΩ (internal resistance of the voltmeter)- t = 4.00 s (time)Substituting the given values into the formula, we have:C=−4.003.40×106·ln(3.0012.00−1)Simplifying this expression, we can evaluate it using a calculator to find the value of C.(b) The time constant (τ) of the circuit is given by the product of the resistance (R) and the capacitance (C):τ=R·CSubstituting the values of R and C, we can calculate the time constant τ:τ=3.40×106·CNow, using the value of C obtained in part (a), we can compute the time constant τ.

A capacitor is charged to a potential of

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