Question on obtaining the ODE

RC circuits :


Why is it that the current flows through the capacitor and the resistor when the voltage source is disconnected. Shouldn't it flow from the capacitor to the resistor since the voltage source is disconnected ?

RL circuits :

Edit1 Why is it that the current flows from the inductor to the resistor here, but in the RC circuit it flows through the resistor and inductor. Also why is that the polarity of the inductor becomes opposite of that of the resistor when the voltage source is switched off.

  • 1
    \$\begingroup\$ Ode (from Ancient Greek: ᾠδή ōidē) is a type of lyrical stanza. ODE to a passive circuit - How charged the flow of particles that define the current's swell, how plates apart can separate the voltages that dwell but if perchance attraction lays in fields of guassian lay where coils of wire resist the flow expressed by volts persay. \$\endgroup\$ Commented Jul 27, 2015 at 17:49
  • \$\begingroup\$ The arrows are there just to indicate the reference direction so you know when to use a positive or negative sign. \$\endgroup\$
    – Curd
    Commented Jul 28, 2015 at 6:32

3 Answers 3


You can define the direction of a current however you want. In the first diagram, \$i_C\$ will be negative if the capacitor voltage is positive at the start.

Similarly, you can define the polarity of a voltage however you want. In the second diagram, to satisfy KVL, \$v_L\$ must equal \$-v_R\$.

The currents in the first diagram make sense when a voltage source is applied to the capacitor and resistor. As the capacitor charges up, \$i_C\$ and \$i_R\$ will both be positive. When the voltage source is removed, \$i_C\$ becomes negative as the capacitor discharges.

The second diagram make less sense. When a voltage (or current) source is applied, the positive ends of \$v_L\$ and \$v_R\$ will be away from ground. When the source is removed, \$v_L\$ goes negative to maintain the inductor current, which means \$v_R\$ must also be negative. So I think the inductor and resistor should be shown with the same polarity, and the current should probably go towards the top of the inductor. But again, it's arbitrary -- you can choose whatever polarities you want, as long as you're consistent. The only difference is the negative sign.

EDIT: In response to your comments, I think you're misunderstanding what's going on. Capacitors have a continuous voltage; inductors have a continuous current. When a capacitor is charged up and the source voltage is disconnected, the capacitor current reverses to try to maintain the voltage. When an inductor is "charged" up and the source current is disconnected, the inductor voltage reverses to try to maintain the current.

The equations are:

$$i_C = C \frac {dv_C}{dt}$$ $$v_L = L \frac {di_L}{dt}$$

When a capacitor and a resistor are connected in parallel, you can use KCL:

$$i_C = -i_R$$ $$C \frac {dv}{dt} = -\frac v R$$

Likewise, when an inductor and a resistor are connected in series, you can use KVL:

$$v_L = -v_R$$ $$L \frac {di}{dt} = -iR$$

Solving these equations give you the usual first-order exponential decay. The initial conditions are the capacitor voltage and inductor current at t=0.

Here's a schematic showing what happens at t=0 in the standard example for this kind of problem:


simulate this circuit – Schematic created using CircuitLab

Before t=0, the switches have been closed for a long time. At t=0, the switches open.

  • \$\begingroup\$ What I understood so far is that in the RC case when the voltage source is disconnected the voltage drop across the capacitor drops from the initial value to zero, and for this to happen dv/dt is infinity because it's a jump discontinuity, and that's undefined. For this reason the voltage drop across the capacitor's terminals becomes the initial value and the same current flows through the capacitor's terminals (ic in the picture). This also means that there is current flowing around the circuit starting from the capacitor's positive terminal (ir in the picture). \$\endgroup\$
    – user404266
    Commented Jul 28, 2015 at 3:46
  • \$\begingroup\$ The problem is that when the same logic is applied to the inductor case. The polarities of the inductor and capacitor must be the same, and there will be to currents one through the inductor and one through the resistor, but that's not the case in the picture. \$\endgroup\$
    – user404266
    Commented Jul 28, 2015 at 3:48
  • \$\begingroup\$ I've updated my answer. \$\endgroup\$
    – Adam Haun
    Commented Jul 28, 2015 at 5:13

enter image description here

The capacitor eventually acquires the potential of the battery,so it flows from +ve terminal of cap to -ve terminal.It flows from the capacitor to the resistor as shown in the diagram.In your case there was probably another source(of higher potential) getting on when the first source was disconnected.

RC discharging circuit

Meanwhile an inductor opposes the flow of i through it,so after the source is disconnected the current flows in opposite direction.

enter image description here


With no source, there must be an initial condition to start proceedings. This may be stored energy: \$CV^2/2\$ for a capacitor or \$LI^2/2\$ for an inductor. Conveniently, the initial condition is expressed as the stored voltage across a capacitor or initial current flowing through an inductor.


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