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Physically what is happening here?

RL My thoughts on this are that the inductor supplies a certain current, and current through a larger resistor dissipates more power.

RC This one I'm not so sure about. I would say the same about voltage, but I think the current caused by this voltage will be proportional to the resistance values and so will the power dissipated by this. Somet

schematic

simulate this circuit – Schematic created using CircuitLab The switch opens after being closed for a very long time. hing isn't right conceptually, can someone help me clear this up?

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  • \$\begingroup\$ Are you talking about when the switch opens after being closed for a certain length of time? \$\endgroup\$ – Andy aka May 30 '13 at 7:17
  • \$\begingroup\$ yep, that is correct \$\endgroup\$ – Brandon May 30 '13 at 7:19
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but I think the current caused by this voltage will be proportional to the resistance values and so will the power dissipated by this

Let's work it out and see. We assume that the source is disconnected at t=0.

For RL and RC circuits with initial energy, i.e., there is an initial current through the inductor or an initial voltage across the capacitor, the current is given by:

\$i(t) = i_0 \ e^{-t/\tau} \$

For the RL circuit:

\$\tau = L/R, \ i_0 = i_L(0) \$

For the RC circuit:

\$\tau = RC, \ i_0 = v_C(0) / R \$

Now, let's calculate the power associated with the resistor R:

\$p_R(t) = i^2(t) R = i^2_0\ e^{-2t/\tau} \$

For the RL circuit:

\$p_R(t) = i^2_L(0)\ e^{-2t/\tau}\ R\$

As we expect for the RL circuit, the power is proportional to the resistance R.

For the RC circuit:

\$p_R(t) = \dfrac{v^2_C(0)}{R^2}\ e^{-2t/\tau}\ R = \dfrac{v^2_C(0)}{R}\ e^{-2t/\tau}\$

So, for the RC circuit, the power is inversely proportional to the resistance R.

How to intuit this without working through the math? Note that the larger the power, the sooner the initial energy is dissipated, i.e., the smaller the time constant.

Now, note the formula for the time constant. For the RL circuit, a larger R gives a smaller time constant while, for the RC circuit, a smaller R gives a smaller time constant.

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  • \$\begingroup\$ My current understanding (with your help ofc) and intuition is that capacitors supply a certain voltage that decays exponentially, a smaller resistor will allow more current and thus more power dissipated in the resistor. \$\endgroup\$ – Brandon May 30 '13 at 19:16
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For the R-L circuit you are correct. When the switch opens, the current previously set-up in the inductor will then flow through the resistor until all the energy previously stored in the inductor has drained away. So if the inductor current were 10A initially and the resistor were 10ohms, there would be a peak voltage of 100V that exponentially decays to zero over time. The time would theoretically be infinity but for practical reasons you can approximate this to 5*L/R. (EDIT corrected error)

Because the initial current into the resistance is determined by the coil current when the switch opens, I^2 R losses means higher power with a higher resistance but remember there is only a finite amount of energy stored in the inductor and so for a higher resistance this power lasts a shorter length of time. Remember the time constant is L/R and bigger R makes the time constant smaller.

For the R-C circuit, the cap will have been previously charged to the voltage of Vs. When the switch opens, the voltage will decay exponentially to zero. Of course the resistor current will have already been flowing due to Vs and after the switch opens this will decay to zero and the time taken is approximately 5*R*C.

enter image description here

The power into the resistor, when the switch opens is determined by the charge voltage and this power is V^2/R so, higher power comes when R is lower. Again there is only a finite amount of energy stored in the capacitor and smaller R means the time constant (C*R) is smaller and this power dissipates more quickly for smaller values of R.

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  • \$\begingroup\$ The time constant for R-L circuits is L/R not L*R (since Henry = Ohm * s) \$\endgroup\$ – GummiV May 30 '13 at 8:27
  • \$\begingroup\$ oops my bad - corrected!!! \$\endgroup\$ – Andy aka May 30 '13 at 8:48

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