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Here is an arbitrary circuit:

enter image description here

After the capacitor is nearly charged, the battery is disconnected. How do I find the power dissipated in each resistor? I know that if the resistors are connected in parallel, then the power dissipated in each is proportional to each resistance because \$P = I^2 R\$ and current in constant when connected in series. But now that the voltage is constant instead of current, is the power dissipated in each resistor inversely proportional to its resistance? Thanks in advance for answering!

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  • \$\begingroup\$ P ="I * V" or "I^2 * R", not "I^R". You should be able to use P=IV and V=IR to come up with an equation relating V, P and R from these \$\endgroup\$
    – BeB00
    Feb 27, 2023 at 2:59
  • \$\begingroup\$ P=V*V/R. Since V keeps changing after disconnecting the battery, P will also change with time. Are you looking to get a plot of the power dissipation with time or do you want an equation of the same? because it will not be a single number answer. \$\endgroup\$
    – sai
    Feb 27, 2023 at 11:39

5 Answers 5

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Well, first of all the capacitor is instandly charged to the battery voltage. So, when discharging we get:

$$\text{V}_\text{C}\left(t\right)=\text{V}_0\exp\left(-\frac{t}{\text{RC}}\right)\tag1$$

Now, in your case we have a parallel combination of resistors, so:

$$\text{R}=\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}\tag2$$

And, in your special case we have \$\text{R}_1=\text{R}_2:=\text{R}_0\$ so we get:

$$\text{R}=\frac{\text{R}_0}{2}\tag3$$

So, we get:

$$\text{V}_\text{C}\left(t\right)=\text{V}_0\exp\left(-\frac{2t}{\text{R}_0\text{C}}\right)\tag4$$

And the power in one resistor is thus given by:

$$\text{P}_{\text{R}_0}\left(t\right)=\frac{\text{V}_\text{C}^2\left(t\right)}{\text{R}_0}=\frac{1}{\text{R}_0}\cdot\left(\text{V}_0\exp\left(-\frac{2t}{\text{R}_0\text{C}}\right)\right)^2=\frac{\text{V}_0^2}{\text{R}_0}\cdot\exp\left(-\frac{4t}{\text{R}_0\text{C}}\right)\tag5$$

And so the total energy is given by:

$$\text{E}_{\text{R}_0}:=\int\limits_\mathbb{R}\text{P}_{\text{R}_0}\left(t\right)\space\text{d}t=\frac{\text{V}_0^2}{\text{R}_0}\int\limits_0^\infty\exp\left(-\frac{4t}{\text{R}_0\text{C}}\right)\space\text{d}t=\frac{\text{R}_0\text{C}\text{V}_0^2}{4\text{R}_0}=\frac{\text{C}\text{V}_0^2}{4}\tag6$$

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I believe you misused the word "constant" in your question, when probably you meant "same" or "equal". Current travelling through series connected devices is not necessarily "constant", but it certainly is "equal". Current through parallel-connected devices is not necessarily equal, but the voltage across them all is. Not constant, but equal.

In your circuit here, nothing is constant immediately after you disconnect the battery, the voltage across all three devices (resistors and capacitor) are equal, but certainly not constant.

The voltage across them all decays exponentially as the capacitor discharges via the two resistors. Initially, immediately after battery removal, both resistors have 5V across them, so the power dissipated in each is:

$$ P = \frac{V^2}{R} = \frac{5^2}{4700} = 5.3mW$$

Alternatively, you can calculate the current through the resistors, and use \$I^2R\$. By virtue of their identical resistance, and equal voltage across them, current through them will also be equal, as Ohm's law tells us:

$$ \begin{aligned} I &= \frac{V}{R} = \frac{5}{4700} = 1.06mA \\ \\ P &= I^2R = 0.00106^2 \times 4700 = 5.3mW \end{aligned} $$ Same result, of course. That's the initial power, the number of Joules of energy per second each resistor receives from the capacitor, but since the voltage is decaying exponentially now, that power will rapidly diminish to zero over time.

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  • \$\begingroup\$ This does not answer the question that was asked. \$\endgroup\$
    – AnalogKid
    Feb 27, 2023 at 17:04
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The power dissipated by a resistor can be stated as \${I^2R}\$ or \${V^2/R}\$

In this case you've got the resistors in parallel across the capacitor, initially charged to a known voltage. For the values given, each resistor will dissipate the same power, but the level will drop logarithmically as the capacitor discharges.

The capacitor voltage time constant can be worked out using standard analysis and will be \$RC/2\$ (The two is because you have two resistors in parallel, so the effective resistance across the capacitor is \${R/2}\$).

If you want the total energy dissipated in each resistor, given that the initial energy on the capacitor is \$CV^2/2\$, and it is equally dissipated across the two resistors, then the energy dissipated in each resistor is \$CV^2/4\$

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  • \$\begingroup\$ That does not answer the question - the total power dissipated after the battery is disconnected. Jan did a nice job showing the derivation of the correct equation. \$\endgroup\$
    – AnalogKid
    Feb 27, 2023 at 17:03
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    \$\begingroup\$ Well the question asks about power, and I've given the formula for resistor power given the voltage. If the questioner wanted the total energy dissipated, then I would have been happy to provide the derivation for that as well. But they asked about power... \$\endgroup\$
    – colintd
    Feb 27, 2023 at 17:25
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"But now that the voltage is constant"? Why would the voltage be constant without the battery connected? It does not make a lot of sense to talk about dissipated power since the resistor has thermal capacity and the actual dissipation in the form of radiated heat happens with some latency. The dissipated additional energy after disconnect is what is stored in the capacitor, \$\frac12 \mathrm C\,\mathrm U^2\$, and it will be distributed half and half.

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The charge of a capacitor will equal battery voltage. The capacitor will not discharge until the voltage drops. When the battery is disconnected, the voltage source comes from the capacitor. The initial power consumption of the resistors can be found with ohms law. When multiple resistors of equal value are connected in parallel rt= r divided by the number of resistors. Once rt is determined, the rc time constant can be used.

Checkout the following: https://www.electronics-tutorials.ws/rc/rc_2.html

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  • \$\begingroup\$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. \$\endgroup\$
    – Community Bot
    Feb 27, 2023 at 12:57

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