In the circuit of Figure P 28.32 , the switch S has been open for a long time. It is then sudden

In the circuit of Figure P 28.32 , the switch S has been...

In the circuit of Figure $\mathrm{P} 28.32$ , the switch $\mathrm{S}$ has been open for a long time. It is then suddenly closed. Determine the time constant (a) before the switch is closed and (b) after the switch is closed. (c) Let the switch be closed at $t=0 .$ Determine the current in the switch as a function of time.

Key Concepts

Time Constant

The time constant is a measure of how quickly a circuit responds to changes. In first-order circuits, it is defined as the product of the resistance and the capacitance (or the inverse product for inductive circuits), and it quantifies the rate at which voltage or current exponentially decays or rises towards its steady state following a disturbance such as a switching action.

Transient Analysis

Transient analysis involves examining the behavior of circuit variables immediately after a sudden change, such as closing or opening a switch. This analysis typically uses differential equations to capture how the circuit response evolves over time from an initial condition toward a new steady-state configuration.

Steady-State Condition

In the context of circuits, the steady-state condition refers to the behavior of the circuit after a long time has passed following any switching action, when all transient effects have decayed. At steady state, capacitors are treated as open circuits, and inductors as short circuits, greatly simplifying the analysis.

Switching Transients

Switching transients occur when a circuit's configuration is abruptly changed, for instance by closing a switch. This sudden modification leads to temporary responses characterized by non-instantaneous changes in voltages and currents, which typically follow an exponential behavior determined by the circuit’s time constant until a new equilibrium is reached.

Transcript

00:01 So in this problem, we have an rc circuit, and we're first going to look at the scenario where we have this switch that's open, and the circuit's been on a long time.

00:10 So what that means when we have a circuit that's been open a long time like this, basically means that the voltage across this capacitor here will be equal to the voltage given by the battery source.

00:22 So we first want to calculate the time constant before we close the switch.

00:26 And what that's going to be is the resistance times the capacitance.

00:32 Now for the resistance, we have to take into account both the 50 kilo -oom and 100 -kil -oom resistor, and since they're in series, their equivalent resistance is just going to be the sum of those two resistances, just 150 -kil -ooms.

00:49 So now when we solve for tau, the time constant, we'll get a value of 1 .5 seconds.

01:01 So now looking at the circuit again, if we close this switch, we want to determine the time constant after closing the switch.

01:11 So what closing the switch does is it creates a short and essentially cuts off the capacitor from the battery.

01:20 So now we have the same time constant equation, but we're just going to use a different resistance.

01:27 So once we close the switch, this capacitor is going to discharge only through this 100 kilo.

01:33 Own resistor.

01:35 So the r that we'll use in this case is just 100 kilomes.

01:40 And we go ahead and do that calculation, we'll find that the time constant in this case is one second.

01:48 So now what we want to do is we want to function for the current that's passing through this switch, so this middle branch here, after we close the switch.

02:00 So what we first need to do is we need to take into account the current due to the battery.

02:05 So coming on this loop, this side of the switch...

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