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I've been given the circuit below.

enter image description here

For \$t>0\$ the switch is closed.

I'm asked to find the charging time for the capacitor, but I'm not so sure how it is charged through which resistors?

I looked up the answer, and it says that the charging time is given by:

\$\tau=(R_1+R_2)\cdot C\$

They justify this by saying you can form a Thevenin circuit with the 1mA source and the \$R_1\$ resistor. Then \$R_1\$ and \$R_2\$ are in series.

But that seems very non-intuitive to me, and it just comes out of nowhere. It's almost as if you have to know, that you can do that.

Can anyone show me another way of solving this problem, cause this solution just seems like some sort of cheap trick.

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    \$\begingroup\$ did you already try to use a current divider approach? There is an example on wikipedia ( en.wikipedia.org/wiki/Current_divider#Example:_RC_combination ) which does the math for a simpler case(R_2=0) but it should be rather easy to extend the existing solution to fit your case. \$\endgroup\$ Dec 28 '19 at 19:17
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    \$\begingroup\$ These types of exercises are setup to encourage you to learn the cheap tricks. Is this a college course question? You will find it difficult to get through without learning the cheap tricks. \$\endgroup\$
    – Mattman944
    Dec 28 '19 at 19:29
  • \$\begingroup\$ The hard way: KCL. Lots of equations, lots of unknowns, algebra. \$\endgroup\$
    – Mattman944
    Dec 28 '19 at 19:31
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    \$\begingroup\$ Shouldn't it be "For t>0 the switch is open."? \$\endgroup\$
    – jDAQ
    Dec 28 '19 at 19:32
  • \$\begingroup\$ Solve all 4 Maxwell equations! Kidding... Next to Mattman944's suggestion using nodal analysis (KCL), you can also use mesh analysis (KVL). But for this circuit it is the harder way. Better search internet and get familiar with source transformation (Norton's and Thévenin's theorems). It will be still a 'trick', but it will safe you time doing an exam. \$\endgroup\$
    – Huisman
    Dec 28 '19 at 20:33
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Well, assuming that the circuit is at rest when \$t<0\$ (that is true because when \$t<0\$ the source is shorted).

Using Laplace transform we can write:

$$\text{V}_\text{in}\left(\text{s}\right)=\text{I}_\text{in}\left(\text{s}\right)\cdot\text{R}_\text{in}\left(\text{s}\right)\tag1$$

Now, we know that:

$$\text{I}_\text{in}\left(\text{s}\right)=\mathcal{L}_t\left[\hat{\text{i}}\right]_{\left(\text{s}\right)}=\frac{\hat{\text{i}}}{\text{s}}\tag2$$

The current through the capacitor is given by the current divider formula:

$$\text{I}_\text{C}\left(\text{s}\right)=\frac{\text{R}_1}{\text{R}_1+\text{R}_2+\frac{1}{\text{sC}}}\cdot\text{I}_\text{in}\left(\text{s}\right)=\frac{\text{R}_1}{\text{R}_1+\text{R}_2+\frac{1}{\text{sC}}}\cdot\frac{\hat{\text{i}}}{\text{s}}=\frac{\hat{\text{i}}\text{R}_1}{\frac{1}{\text{C}}+\left(\text{R}_1+\text{R}_2\right)\text{s}}\tag3$$

Using inverse Laplace transform we get:

$$\text{i}_\text{C}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\frac{\hat{\text{i}}\text{R}_1}{\frac{1}{\text{C}}+\left(\text{R}_1+\text{R}_2\right)\text{s}}\right]_{\left(t\right)}=\frac{\hat{\text{i}}\text{R}_1}{\text{R}_1+\text{R}_2}\cdot\exp\left(-\frac{t}{\text{C}\left(\text{R}_1+\text{R}_2\right)}\right)\tag4$$

Using the voltage-current relation in a capacitor \$\text{i}_\text{c}\left(t\right)=\text{C}\cdot\text{v}_\text{c}'\left(t\right)\$, we get:

$$\text{v}_\text{C}\left(t\right)=\hat{\text{i}}\text{R}_1\left(1-\exp\left(-\frac{t}{\text{C}\left(\text{R}_1+\text{R}_2\right)}\right)\right)\tag5$$

Generally, we call the number inside the \$\exp\$-function the time constant of the circuit \$\tau\$. In this circuit we have:

$$\tau=\text{C}\left(\text{R}_1+\text{R}_2\right)\tag6$$

We also say that a capacitor is charged when \$5\tau\$ seconds are passed.

So the charge up time is given by:

$$5\tau=5\text{C}\left(\text{R}_1+\text{R}_2\right)\tag7$$

The voltage across the capacitor is then given by:

$$\text{v}_\text{C}\left(5\tau\right)=\hat{\text{i}}\text{R}_1\left(1-\exp\left(-\frac{5\text{C}\left(\text{R}_1+\text{R}_2\right)}{\text{C}\left(\text{R}_1+\text{R}_2\right)}\right)\right)=$$ $$\hat{\text{i}}\text{R}_1\left(1-\frac{1}{e^5}\right)\approx0.993262\hat{\text{i}}\text{R}_1\tag8$$

The current trough the capacitor is then given by:

$$\text{i}_\text{C}\left(5\tau\right)=\frac{\hat{\text{i}}\text{R}_1}{\text{R}_1+\text{R}_2}\cdot\exp\left(-\frac{5\text{C}\left(\text{R}_1+\text{R}_2\right)}{\text{C}\left(\text{R}_1+\text{R}_2\right)}\right)=$$ $$\frac{\hat{\text{i}}\text{R}_1}{e^5\left(\text{R}_1+\text{R}_2\right)}\approx0.00673795\cdot\frac{\hat{\text{i}}\text{R}_1}{\text{R}_1+\text{R}_2}\tag9$$

In your circuit we get:

  • $$5\tau=5\cdot10\cdot10^{-6}\left(10000+20000\right)=\frac{3}{2}=1.5\space\text{s}\tag{10}$$
  • $$\text{i}_\text{C}\left(5\tau\right)=\frac{1}{3000e^5}\approx2.24598\space\mu\text{A}\tag{11}$$
  • $$\text{v}_\text{C}\left(5\tau\right)=10\left(1-\frac{1}{e^5}\right)\approx9.93262\space\text{V}\tag{12}$$
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(considering this might be homework)

Try using some Norton->Thévenin transform on the current supply and one of the resistors, it will make your life way easier and the solution will be pretty much that.

But, considering you find that counter intuitive. Think of what happens when the capacitor completely charges. If the capacitor is charged that means no current goes through the capacitor branch, so all the current \$ I \$ goes through \$ R_1\$, from that you can figure out to what voltage the capacitor will charge (\$ V_C = I R_1\$). Now you can use KCL and KVL to have, $$ I = i_{R1}(t) + i_{R2}(t), $$ $$ R_1i_{R1}(t) = R_2i_{R2}(t) + V_{C}(t), $$ $$ V_{C}(t) = \dfrac{1}{C}\int^t_0{i_{R2}(t)dt}. $$ That should be enough to solve it.

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  • \$\begingroup\$ You're suggesting something OP already knows! So, either convince OP it does not come out of nowhere or show different approaches OP is asking for. (I'd go first Thevenin, you're right it is easiest approach) \$\endgroup\$
    – Huisman
    Dec 28 '19 at 20:09
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But that seems very non-intuitive to me, and it just comes out of nowhere. It's almost as if you have to know, that you can do that.

That's absolutely fine, don't worry; it does take time to see it straight off.

What about charging? Until the switch opens, the capacitor cannot charge because it's shorted by the switch where all the current flows through. Notice this is strictly tied to an ideal circuit i.e. real world won't be this way. But it doesn't matter at this stage.

When the switch opens, no current flows in that branch so you might even remove it from the schematics. You get \$I_s\$ with \$R_1\$ in parallel: this resembles a real current source (ideal current source with a resistance in parallel), so it's safely equivalent to (can be formally proven with Thevenin's theorem):

schematic

simulate this circuit – Schematic created using CircuitLab

Now it should be clear why the resistance seen is the series of the two resistors. Is that extra step really needed? Nope, but it's helpful generally. The best way is to turn off all the sources (i.e. open the current source), substitute a voltage source \$v_x\$ in place of \$C\$ and given \$v_x\$ calculate \$i_x\$, which is the current the generator provides, by e.g. Kirchhoff's laws. You'll find the same series, of course.

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