Where am I wrong in my derivation of RC charging circuit equation?

Question

I'm trying to derive the RC charging circuit equation for capacitor voltage \$V_C\$ and charging time \$t\$. Here's my solution,

We'll start by assuming the battery, resistor and capacitor are connected in series.

• KVL states that sum of all voltages in series is zero. This gives us \$V_B+V_R+V_C=0\$.
• KCL states that current is equal everywhere in a node. We only have a single node where everything is connected in series. This gives us \$I=I_C=I_R=I_B\$.

Substituting the values for \$I_C\$ and \$I_R\$ in KCL we get, \$C\frac{dV_C}{dt}=\frac{V_R}{R}\$ where \$V_C\$ is the voltage across the capacitor and \$V_R\$ is the voltage across the resistor. Although from KVL above, we know that \$V_R=-V_B-V_C\$ or \$V_R=-(V_B+V_C)\$ if we take minus common. Now, Substituting the value for \$V_R\$ in our previous equation we get, \$C\frac{dV_C}{dt}=\frac{-(V_B+V_C)}{R}\$. Rearranging this would yield, \$\frac{1}{V_B+V_C}dV_C=-\frac{1dt}{RC}\$. Taking integral on both sides now yields, \$log_e(V_B+V_C)=-\frac{t}{RC}\$.

Note this equation \$log_e(V_B+V_C)=-\frac{t}{RC}\$. I'll use this to derive time \$t\$.

Taking exponential on both sides yields, \$e^{log_e(V_B+V_C)}=e^{-\frac{t}{RC}}\$. This simplifies into, \$V_B+V_C=e^{-\frac{t}{RC}}\$ which upon rearranging yields, $$V_C=-V_B+e^{-\frac{t}{RC}}$$.

Solving for initial condition's where \$t=0\$ and \$V_C=V_{C0}\$ in our solution above gives us \$V_{C0}=-V_B+e^{-\frac{0}{RC}}\$. This simplifies to, \$V_{C0}=-V_B+e^0\$ and then \$V_{C0}=-V_B+1\$. Putting \$V_B\$ on the left hand side yields, \$V_B=-V_{C0}+1\$. Now If I substitute \$V_B\$ into the equation above \$V_C=-(-V_{C0}+1)+e^{-\frac{t}{RC}}\$. Rearranging this yields, $$V_C=V_{C0}+(e^{-\frac{t}{RC}}-1)$$

However the standard equation is \$V_C=V_B(1-e^{-\frac{t}{RC}})\$ and I'm not sure what I did wrong.

Now deriving time from \$log_e(V_B+V_C)=-\frac{t}{RC}\$. We can arrange this to have \$t\$ on the left hand side, this results in \$-t=RC*log_e(V_B+V_C)\$. Multiplying \$-1\$ on both sides yields \$t=-1*RC*log_e(V_B+V_C)\$. We know \$-1*log(x)=log(\frac{1}{x})\$. Therefore our solution for time \$t\$ becomes $$t=RC*log_e(\frac{1}{V_B+V_C})$$.

However the standard equation for time is \$t=RC*log_e(\frac{V_B}{V_B+V_C})\$ and I'm not sure where I went wrong.

update: clarifying integral

This is how I solved the integral \$C\frac{dV_C}{dt}=\frac{-(V_B+V_C)}{R}\$.

1. We can divide by \$(V_B+V_C)\$ on both sides. This results in, \$C\frac{dV_C}{(V_B+V_C)dt}=\frac{-1}{R}\$.
2. We can multiply by \$dt\$ on both sides. This results in, \$C\frac{dV_C}{(V_B+V_C)}=\frac{-1}{R}*dt\$.
3. We can divide by \$C\$ on both sides. This results in, \$\frac{dV_C}{(V_B+V_C)}=\frac{-1}{RC}*dt\$.
4. Now we can take the integral on both sides, \$\int\frac{dV_C}{(V_B+V_C)}=\int\frac{-1}{RC}*dt\$.
5. On the left hand side we have \$\int\frac{1}{V_B+V_C}dV_C\$. We know \$\frac{1}{x}=log_e(x)\$, therefore our left hand solution becomes \$log_{e}(V_B+V_C)\$.
6. On the right hand side we have \$\int\frac{-1}{RC}dt\$. \$\frac{-1}{RC}\$ is constant so we can pull it outside and thus our right hand integral becomes \$\frac{-1}{RC}\int1dt\$. This simply results in, \$\frac{-1}{RC}*t\$ and then \$\frac{-t}{RC}\$.
7. Combining left hand side and right hand side solutions from step 5 and step 6, we end up with \$log_{e}(V_B+V_C)=\frac{-t}{RC}\$

Is this wrong?

Answer 1

KVL states that sum of all voltages in series is zero.

There is a constraint: The voltages of each element must be measured with the same orientation. The diagram aligns the voltages with the direction of the Kirchhoff path. They can be aligned opposite, but must be consistent.

Some consider voltage sources as positive with the passive element voltages as negative. So $$+V_1-V_{R1}-V_{R2}=0$$

^{simulate this circuit – Schematic created using CircuitLab}

KCL states that current is equal everywhere in a node.

This is not right. KCL is:

"The sum of the currents into a node equals zero."

We only have a single node where everything is connected in series.

This also is not right. A series circuit of three elements has three nodes with two branches for each node.

KCL must be applied twice to each of node a and node b showing the the element currents are equal.

For example at node a, $$ I_{a1}+I_{a2}=0 $$ Substituting element currents $$ I_{B}-I_{R}=0 $$ showing that $$ I_{B}=I_{R} $$

This is quite pedantic, so recognizing that elements in series must have the same current is much easier.

^{simulate this circuit}

Your analysis:

Substituting the values for \$I_C\$ and \$I_R\$ in KCL we get, \$C\frac{dV}C{dt}=\frac{V_R}{R}\$

This is correct.

from KVL above, we know that \$V_R=-V_B-V_C\$ or \$V_R=-(V_B+V_C)\$

Here is the first error. It should be \$V_R=V_B-V_C\$

The differential equation should look like this:

$$ RC\frac{dv_C}{dt}+(v_C-v_B)=0 $$

So this is the second error. A change of variable must be introduced. You are attempting this but need to be careful. Also be clear that the shape of \$v_B\$ must be known. In this case it is a step of amplitude \$V_B\$, which is a constant for \$t\ge 0\$. So let \$x=(v_C-V_B)\$. Then $$ RC\frac{dx}{dt}+x=0 $$

The solution to this is $$ x=X_I e^{\frac{-t}{RC}} $$ where \$X_I\$ is the initial value of x.

Back substituting: $$ (v_C-V_B)=-V_Be^{\frac{-t}{RC}} $$

You should be able to take it from here. When solving problems of this type be certain to perform change of variable correctly.

Answer 2

The main thing is you lost track of the polarity of your voltages. A diagram can be very helpful in keeping track of this. From your first point about KVL — \$V_B + V_R + V_C = 0\$ — you have:

^{simulate this circuit – Schematic created using CircuitLab}

Obviously, at least one of the voltages needs to be negative in order for that to work. The "standard equation" assumes that the polarities of the resistor and capacitor are reversed from this. In other words, they simply write \$V_B = V_R + V_C\$ and go from there.

Also, your integration step is wrong. You're integrating with respect to \$t\$, but you ended up losing track of the \$dV_C\$ somewhere along the way.

Answer 3

The error is introduced in the integration, because the resulting equation has voltage unit on left side and no unit on right side.