The switch S is initially open and the capacitor voltage is 75 V. The switch is
ID: 1322035 • Letter: T
Question
The switch S is initially open and the capacitor voltage is 75 V. The switch is then closed at time t = 0. How long after closing the switch will the current in the resistor be 12.0 microAmps?
Please I am looking to learn so if you could explain your steps in detail, why and how you did things it would be greatly appreciated! Thank you for your time
The switch S is initially open and the capacitor voltage is 75 V. The switch is then closed at time t = 0. How long after closing the switch will the current in the resistor be 12.0 microAmps? Please I am looking to learn so if you could explain your steps in detail, why and how you did things it would be greatly appreciated! Thank you for your timeExplanation / Answer
The voltage across the resistor, which will define the current through it by Ohm's Law, can be measured as Vr(t) = Vs*e^(-t/(R*C)), where Vs is the source voltage (500V here), R is the resistance (200 ohms here), and C is the capacitance in a simple RC circuit (12uF here). This is because the voltage across the capacitor increases as it is charged with a current. Since our source is constant, the total voltage within the circuit must also remain the same. Thus, as the capacitor increases in voltage, the resistor experiences less voltage, which means the current in the circuit must also lower with this decrease in voltage. As the current decreases, the capacitor will charge slower and slower in an exponential arc. In reality, the capacitor is experiencing a voltage of Vc(t) = Vs*(1 - e^(-t/R*C)). Note that if you add Vc(t) + Vr(t) together, you are left with only Vs.
Back to the original formula for the resistor voltage, Vr(t) = Vs*e^(-t/RC) = Vs*e^(-t/T). The time constant, T, is an expression of an arbitrary amount of time required before certain levels of voltages are experienced and equal to R*C, so when one time constant has passed, t = R*C. At t = T, roughly 63.25% of the voltage will be in the capcaitor, 36.75% in the resistor. At t = 2*T, the values become 86.5% and 13.5% respectively. When t = T, Vr(t) = Vs*e^(-R*C/R*C) = Vs*e^-1. Thus Vr(t) = 500/e = 183.94V. By ohm's law, I = V/R, so I = 183.94V/200ohms = 0.9197A, so it is closest to {D}.
At that instant, the cap has no charge and will take all it current it can get. So, at that instant, the cap has no effect and may be replaced by a short (no charge (0V) and will take all the current). Then it is a simple resistive circuit. It is key that you understand that. That will get you 1, 2, 3.
4 is the opposite. The switch has been open long time. There is no current in the cap and it is fully charged. The cap is replaced by an open circuit. No current in R3 and has the voltage of R2.
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