# Circuit short circuit capacitor question confusion

Here's the question:

Determine the value of the current labeled $i$ and the voltage labeled $v$ at $t=0+$, $t=1.5ms$, and $t = 3.0ms$.

The circuit is as follows:

simulate this circuit – Schematic created using CircuitLab

Relevant equations:

Ohm's law: $v = iR$

Decay in source free RC Circuit $v(t) = v(0) e^{-t/RC}$

The attempt at a solution:

$i(0-) = 0.1 A.$

Therefore,

$v$ across 200 ohm: $200 * 0.1 = 20 V$

Since no current flows through capacitor with constant dc current,

$v$ across capacitor: $20 V @ t < 0$

$v$ across capacitor cannot change immediately as it would require infinite power so $v$ across capacitor @ $t = 0$ is $20V$

Here's where the confusion is: Should I include the 200 ohm resistor in my RC circuit? Why or why not? Also, will my circuit be a source free RC circuit?

Also, what will be the current i through the 200 ohm resistor when t>0?

My last question is a theoretical one: Can a current through a resistor change immediately at one instant of time?

• Use the laplace transform to treat the capacitor as a complex impedance, apply the usual circuit analysis to the combination of impedances as if they were simple resistances. Then evaluate the time domain response to stimulus. – Chris Stratton Oct 22 '12 at 16:13
• Hint: It's a trick question: When the switch is closed, how much voltage is there across the 200-ohm resistor? – Dave Tweed Oct 22 '12 at 16:15
• @ChrisStratton I haven't learned complex impedance yet at my college so I don't have many idea what you are talking about. Please check out my edit. – Prabhpreet Oct 22 '12 at 16:17
• @ChrisStratton: Yes, it is. The diagram clearly shows the switch closure happening at t=0, and the evaluation times are all for t>0. Therefore, all the current in both loops is flowing through the switch, and there's no current flowing in the resistor. – Dave Tweed Oct 22 '12 at 16:29
• -1 for ridiculously large but blurry images and quite dark white level. You don't need this size image in the first place, and considering the blurriness, much smaller images would have carried the same information. – Olin Lathrop Oct 22 '12 at 17:16

The voltage across a capacitor discharging into a fixed resistance decays exponentially. The time constant is RC, where C is the capacitance, and R is the resistance between the terminals of the resistor.

$$V(t)=V(0)e^{\frac{-t}{RC}} \\$$

You're right that the capacitor starts at 20V, because that's the voltage across the 200 ohm resistor after the capacitor is charged. So you know V(0), and you know C. All you're missing is R. To analyze that properly, think of the switch as a resistor of 0 ohms when closed, and don't worry about the current source. The parallel combination of a 200 ohm resistor and a 0 ohm resistor is 0 ohms. The series combination of a 50 ohm resistor and a 0 ohm resistor is 50 ohms. So the 50 ohm resistor is the only one that matters when determining your discharge constant.

The current through the 200 ohm resistor depends on the voltage across the 200 ohm resistor. The voltage across the 200 ohm resistor is the same as the voltage across the 0 ohm resistor (the closed switch). V=IR, so what's the voltage across that pair of resistors when the switch is closed?

And yes, the voltage across an ideal resistor can change instantaneously. Keep in mind though, there's no such thing as an ideal resistor in the physical world. Everything has capacitance to everything else.

• A short circuit should have no potential across it so the voltage across the 200 ohm resistor should be zero, right? – Prabhpreet Oct 22 '12 at 17:43
• @Prabhpreet Right on. – Stephen Collings Oct 22 '12 at 17:55
• @Prabhpreet: Yes. A short circuit has no potential across it. So the voltage across that 200 Ohm resistor (after t=0+) is zero. – davidcary Oct 22 '12 at 17:57

When solving problems like this, it is helpful to draw the two different circuits that exist before and after $t=0$.

Before, with the switch open, since the circuit is in DC steady state, replace the capacitor with an open circuit and then it is easy to solve for $v_C(0)$, the initial condition.

After, with the switch closed, the current source and 200 $\Omega$ resistor are paralleled with a short circuit and so, from the perspective of the capacitor, can be ignored (a short circuit in parallel with any other circuit elements is equivalent to a short circuit).

So, the circuit after $t=0$ is just the capacitor and the 50 $\Omega$ resistor, a simple RC circuit with initial condition $v_C(0)$.

Also, what will be the current i through the 200 ohm resistor when t>0?

With a short circuit in parallel, the voltage across the resistor is ?V and, using Ohm's law, the current is ?A.

My last question is a theoretical one: Can a current through a resistor change immediately at one instant of time?

Theoretically, within ideal circuit theory, the answer is yes. In reality, the answer is no. for example, physical resistors have associated "parasitic" capacitance and inductance.

• Sorry I cannot understand what you imply by ?V and ?A. Do you mean that it cannot be found. I feel that the potential across the short circuit should be zero hence the potential across 200 ohm resistor being zero. – Prabhpreet Oct 22 '12 at 17:44
• @Prabhpreet, sorry I wasn't clear. The question marks were meant to be prompts for you to fill in the answer. You're correct, the voltage across a short circuit is zero (by definition!) and, since this parallels the resistor, the voltage across the resistor is zero too. Then, by Ohm's law, zero volts across a (non-zero) resistor gives zero current. – Alfred Centauri Oct 22 '12 at 20:23

at t=0 the equivalent circuit is very simple.

Since the switch is ideal, it will have 0V drop circulating all the current source and thus closing the path out of the circuit.

Thus the 20V initial condition discharges with an RC time constant of 20uF*50R=1000 uS.

Did anyone actually guess the answer at 1.5mS and 3mS is 5V(~25%) and 1V(~5%) respectively for the exponential decay from 20V to 0V?

• v @ 1.5 ms ~= 4.5 V, v @ 3ms ~= 1V – Prabhpreet Oct 23 '12 at 4:36

The original post mentions current 'i' etc, but the current 'i' etc is not shown in the original circuit diagram. Also, the question is not clearly defined because for these kinds of problems, it's necessary to state the position of the switch (open or closed) at time t < 0. If assuming the switch is open at t < 0, then the initial voltage of the capacitor Vc(0-) is 20V, which also equals Vc(0), which also equals Vc(0+) because the capacitor voltage is continuous. Assuming the switch closes at t = 0, you end up with a capacitor in parallel with the 50 Ohm resistor, so the time constant RC will be based on C and the 50 Ohm resistor, and you get the capacitor voltage equation as being: Vc(t) = Vc(0)e^(t/50.C), where Vc(0) is 20V. The current ic(0) will be zero because voltage can't change instantaneously when the switch closes at t = 0. So the voltage remains constant at t = 0, in which case the dvc(t)/dt = 0 at t = 0, so zero capacitor current at t = 0. But at t = 0+ (a touch greater than t=0), ic(t) = CdVc(t)/dt = So knowing the Vc(t) equation will yield the capacitor current equation ic(t) for t >= 0+. And the reference direction for the current is in the same direction as the voltage drop across the resistor.....in other words through the capacitor and 'downwards' (starting from the + terminal of the capacitor and ending at the - terminal of the capacitor).

At t < 0, if the switch is open, then the 200 Ohm resistor and the current source simply produce the 20V at the upper node, which drives the initial capacitor voltage to 20V. Once the switch closes at t = 0, the 200 Ohm resistor disappears because it gets short circuited (and taken out of the picture) by the closed switch. It all depends on what the switch is doing at t < 0 and t = 0, which is why you need to properly define the conditions of the problem/question.

• This is unfortunately a (correct) comment rather than an answer. – Kaz Oct 20 '13 at 9:01

Current in an inductor cannot change suddenly (i.e in zero time). Similarly voltage across a capacitor cannot change suddenly.So $Vc(0+) = Vc(0-)$. At $0-$ time capacitor is an open circuit (because after a long duration capacitor cannot charge anymore and takes 0 current). So $Vc(0-)$ is $20V$ which is equal to $Vc(0+)$.

Any transient analysis can be done using $x = A+Be^{-t/timeconstant}$ formula where $x$ can be either voltage or current and A and B are constants which we calculate from known things in the circuit. Comment if you want to know more about how to solve using this method.

• yes i would like to know more. – Prabhpreet Oct 23 '12 at 2:47