Energy Loss Incurred when Charging a Capacitor in a RC Circuit

Physics Tutorial

Energy Loss Incurred when Charging a Capacitor in a RC Circuit


Among their many applications, capacitors are often used as short-term energy storage elements in electronic systems. When a capacitor charges in a simple series RC (resistor-capacitor) circuit, the energy stored in the capacitor increases as it charges and the resistor dissipates energy as the capacitor charging current passes through it. Series resistance is present in any circuit, and is often intentionally placed in capacitor charging circuits to limit the charging current to a safe level. In today’s technology-rich environment with its host of battery-powered hand-held devices, minimizing unnecessary energy loss is a high priority. In light of this, it is important to be able to calculate the dissipative losses in all types of circuits, and keep these losses to a minimum. In this tutorial we calculate the amount of energy dissipated in the resistor contained in a series RC circuit as the capacitor is charged (as a reminder, an ideal capacitor is a reactive element that does not dissipate any energy). Intuition may tell you that the energy loss depends on the value of the resistor. The answer may surprise you.

Sample Problem

Use the following information to answer the question below.

After studying this tutorial you should be able to answer the following question. Approximately how much energy is dissipated in the resistor contained in a series RC circuit constructed with R = 1000 ohms and C = 100 microfarad as the capacitor is charged from 0 volts to 10 volts?

We start with a simple RC circuit consisting of an uncharged ideal capacitor C in series with an ideal resistor R, an ideal switch SW, and an ideal DC voltage source V. When the switch is closed, current begins to flow as the capacitor charges. Since the capacitor is initially uncharged and its voltage cannot change instantaneously, the instantaneous initial current is simply V/R. As the capacitor charges, the voltage across the resistor decreases by KVL (Kirchhoff’s Voltage Law), thereby decreasing the current in the circuit. The current in this type of natural system exhibits an exponentially decaying behavior. This current passing through the resistor causes energy to be dissipated from the circuit. How much energy is dissipated in the resistor as the capacitor is charged to some voltage Vc?

1 millijoule

500 microjoules

25 millijoules

10 millijoules

5 millijoules


Let’s start with a picture of the circuit:

RC Circuit

The initial voltage across the capacitor is 0 volts. The instant the switch is closed, the capacitor voltage is zero and all of V appears across R. The current at t = 0 is therefore V/R. We define this initial current as Ip in which the subscript “P” indicates “Peak.” The current decays exponentially from Ip to the asymptote of 0 amps as the capacitor charges. We also define the time constant τ as the product of the resistance and capacitance, thus τ = RC.

The equation of the current i(t) as a function of time is described as follows:

    \[     i(t) \: = \: I_Pe^{-t/\tau} \]

The next step is to calculate the instantaneous power, P(t), dissipated in the resistor as a function of time. The power P(t) dissipated in the resistor is calculated as

    \[     P(t) \: = \: i^{2}(t)R \: = \: I_P^{2}e^{-2t/\tau}R \: = \: I_P^{2}Re^{-2t/\tau} \]

We now need to integrate the power from t = 0 to t = ∞ to get the total energy, E, dissipated in the resistor. For simplicity, the initial condition will be zero volts on the capacitor, therefore the constant of integration = 0.

    \[     E \: = \: \int_{0}^{\infty} {I_P^{2}R}e^{-2t/\tau} , dt \: = \: {I_P^{2}R}\int_{0}^{\infty} e^{-2t/\tau} , dt \]

To simplify the integration, substitute a = -2/τ.

    \[     E \: = \: {I_P^{2}R}\int_{0}^{\infty} e^{at} , dt \: = \: {I_P^{2}R}\Big[\frac{e^{at}}{a} |_{t=\infty} - \frac{e^{at}}{a} |_{t=0}\Big] \]

Substituting -2/τ for a

    \[     E \: = \: {I_P^{2}R}\Big[\frac{e^\frac{-2t}{\tau}}{-2/\tau} |_{t=\infty} - \frac{e^\frac{-2t}{\tau}}{-2/\tau} |_{t=0}\Big] \]

    \[     E \: = \: {I_P^{2}R}\Big[0 - \frac{-\tau}{2}\Big] \: = \: {I_P^{2}R}\frac{\tau}{2} \: = \: \frac{I_P^{2}R\tau}{2} \]

We can express Ip as V/R, and realizing that the final charged capacitor voltage Vc is equal to the applied voltage V, we can write Ip = Vc/R and substitute this into our energy equation. We can also substitute RC for τ. Doing both of these, we arrive at the following result:

    \[     E \: = \: \frac{I_P^{2}R\tau}{2} \: = \: \frac{1}{2}\Big[\frac{V_C}{R}\Big]^2\Big[R\Big]\Big[RC\Big] \: = \: {\frac{1}{2}}C{V_C}^2 \]

This amazing result shows us that the energy dissipated in the resistor when charging the capacitor is equal to the energy stored in the capacitor! It is also interesting that the energy dissipated in the resistor does not depend on the value of the resistor.

Another way to look at this is that when charging the capacitor in a series RC circuit, half of the energy supplied by the voltage source is dissipated in the resistor and the other half is stored in the capacitor. This basic result holds for any amount of energy stored in the capacitor. In other words, if we had an initial voltage on the capacitor or if we opened the switch before t = ∞ the amount of energy change in the capacitor is equal to the amount of energy dissipated in the resistor. If we look at this intuitively, it makes sense because if we discharge the capacitor, the resistor must dissipate the amount of energy supplied by the capacitor.

Let’s take this result a step further and see how we can apply it to a practical situation in which we store energy supplied by a battery in a capacitor for later use and wish to know what percent of the fully-charged battery’s energy we will waste in the resistor when we charge the capacitor. We will use a large “supercapacitor” that has a very large capacitance to store the energy. Let’s charge a 5 farad supercapacitor from 0 volts to 8 volts through a 10 ohm resistor. The energy stored in the supercapacitor is 0.5(5 F)(8 V)^2 = 160 joules. This is also how much energy is dissipated in the 10 ohm resistor. We will use a 8 volt battery with a capacity of 5 amp-hours to charge the supercapacitor and see what percentage of the fully-charged battery’s energy is dissipated in the resistor. The energy stored in the battery is calculated as E = (8 volts)(5 amp-hours)(3600 seconds/hour) = 144,000 joules. The energy dissipated in the resistor is therefore only about 0.11% of the energy stored in the battery. Another 0.11% of the battery’s energy is stored in the supercapacitor.

Next time we will take our fully-charged supercapacitor and discharge it through a boost switching voltage regulator that provides a constant output voltage to power our system for a short time as the capacitor voltage decays. We then solve a differential equation to calculate how long the system can be powered before the capacitor discharges to the point where it is no longer effective. Stay tuned!

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