Calculating Charge and Time in a Discharging RC Circuit

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In the discussion about calculating charge and time in a discharging RC circuit, the key points focus on determining the charge in the capacitor when the current is 33μA and the time it takes for the current to reach that level. The initial voltage across the capacitor is 80V, with a resistance of 1.5MΩ and a capacitance of 22μF. The user suggests using the formula Q = IR to find the charge and considers applying the exponential decay formula Q = Q0 e^(-t/RC) for calculating time. There is confusion regarding the initial charge Q0 and whether it can be calculated as C multiplied by the initial voltage. The conversation indicates a need for clarity on the circuit diagram and the calculations involved.
Moolisa
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Homework Statement


RC circuit, with a resistor, capacitor and a switch with
ΔVC =80V R=1.5MΩ C=22μF
The switch is initially open. The switch is closed at t=0
(a) What is the charge in the capacitor when the current is 33μA
(b) How times does it take for the current to reach 33μA
2. The attempt at a solution
(a) So I think the capacitor is discharching.
Since ΔVC -IR=0, and ΔVC=Q/C is the charge when I=33μA
Q=IRC?
(b) If part (a) is correct, can I use Q=Q0 e-t/RC and find the time from there?
 
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It looks correct to me. For part b, how do you compute ## Q_o ##? I don't think you showed that.
 
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Charles Link said:
It looks correct to me. For part b, how do you compute ## Q_o ##? I don't think you showed that.
Is it C(80V)?
 
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I'd like to see the circuit.
 
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