Why is resistor used in RC circuit?

Question

Why is the resistor used in RC circuit? Is resistor used to oppose the current flow in RC circuit?

Answer 1

The resistor is indeed being used to control the amount of current flowing in the loop consisting of resistor, capacitor and voltage source.

To understand the role of the resistor in more detail, first you must understand what happens to the voltage across a capacitor, as a current flows through it.

The voltage across a capacitor increases at a rate which is proportional to the current flowing through it. As an example, let's say we have a completely discharged 1μF capacitor. It will have a potential difference of 0V across it. If we then pass 1μA of current through it, the voltage across it will begin to rise, at a rate of 1V each second. After 1 second it will have 1V across it, after 2s that voltage will have risen to 2V, and so on. Given a capacitance \$C\$ and current \$I\$ through that capacitance, the formula for the rate of change of voltage across the capacitor \$\frac{dV}{dt}\$ is:

$$ \frac{dV}{dt} = \frac{I}{C} $$

If I plug in the numbers from above into this formula, we get a rate of change of voltage of:

$$ \frac{dV}{dt} = \frac{I}{C} = \frac{1\mu A}{1\mu F} = 1\frac{V}{s} $$

Importantly, that formula tells you that the amount by which voltage changes, each second, (that's what \$\frac{dV}{dt}\$ means) is proportional to the current \$I\$ flowing. If you double the current, you double the rate at which voltage changes, if you reduce current by a factor of 10, the voltage across the capacitor changes 10 times slower, and so on.

In other words, if we can somehow control that current, we control how fast the capacitor's voltage changes. That's the role of the resistor, which we use to specify how much current will flow, and consequently how fast the capacitor charges (or discharges, for that matter). This has a thousand uses.

Imagine, for example, that you have a circuit that passes some current through a capacitor, and waits for capacitor voltage to exceed, say, 5V. When that happens, it reverses current flow, and waits for the capacitor to discharge back to 0V. Then it reverses current again, and starts to re-charge the capacitor, and so on, again and again. Such a circuit is called a relaxation oscillator, because it switches back and forth between states, charging and discharging the capacitor, over and over.

I'm sure you can see that the time it takes for capacitor voltage to reach 5V, and the time it takes to get back to 0V again will depend on how fast that voltage is rising and falling, which will depend entirely on the amount of current through the capacitor. If we charge and discharge using large currents, frequency of oscillation will be high, if we use small currents, frequency will be much lower. A resistor in this scenario is a perfect way to control how much current flows through the capacitor, and therefore the frequency of oscillation.

Since you refer to "RC circuit", I'll also try to demonstrate how the resistor in there will affect the circuit's behaviour. Let's look at the low-pass RC filter:

^{simulate this circuit – Schematic created using CircuitLab}

Those are three RC filters, all with the same capacitance, but the resistors get progressively larger in value. Each of the filters is presented with the same input signal, a triangular voltage waveform that looks like this:

The three outputs, OUT1 (blue, 10Ω resistor), OUT2 (orange, 100Ω resistor) and OUT3 (tan, 1kΩ resistor) look like this:

As you can see, OUT1 is almost a perfect copy of the input. This is because the 10Ω resistor offers very little resistance to current, and the capacitor is able to charge very quickly, at a rate almost perfectly able to match the input's own rate of change. It can "keep up with the input", so to speak.

The second circuit, with the 100Ω resistor, struggles to keep up. That resistor reduces the amount of current that will flow for some given input potential. Consequently, the input changes at a rate which exceeds the capacitor's ability to "keep up". The capacitor doesn't have time to get all the way to the top or bottom, before the input changes direction. The result is a slightly smaller signal amplitude at OUT2.

OUT3 is the result of even smaller charge/discharge current, due to the even greater resistance R3. It charges and discharges so slowly that it has no hope of ever keeping up with the changing input.

If it's not clear why this is of any benefit, then try looking at what happens when instead of changing resistance, we change the frequency of input signal, keeping the resistance unchanged. I will use the third circuit, with C3 and R3, and I will use a sinusoidal input instead of a triangular one. I will sweep the input frequency from about 200Hz to 1kHz, and plot the input potential \$V_{IN}\$ at IN and output \$V_{OUT3}\$ at OUT3:

It's pretty clear that as input signal frequency rises, output amplitude decreases, which happens because the faster the rate of change of input, the more difficult it is for capacitor voltage to keep up. This filter passes lower frequency signals very well, but attenuates higher frequency ones, which is why it's called a low-pass filter.

As I mentioned before, if we use a lower resistance, we increase capacitor current, and the capacitor could respond more quickly to input changes. This means that the circuit will start to attenuate at a higher frequency. The resistance therefore determines the frequency above which this circuit begins to attenuate. In fact, both resistance and capacitance determine this "cut-off" frequency. Without going deeply into the maths, the formula is:

$$ f = \frac{1}{2\pi RC} $$

That's the frequency where a sinusoidal input will be attenuated to about 70% of the input amplitude.

This is just a contrived example of the resistor's role in an RC filter; there is much, much more to learn about it, which could take days, or months, or (hopefully not) years to grok fully. How long it takes to charge your your information-capacitor with information will depend entirely on the resistance present between sources of information and your information-capacitor.

Answer 2

Basic idea

Two-terminal circuit elements are described by the relationship between their two quantities current and voltage (aka IV curve). If we change one quantity as input and the other as output, we can consider them as converters. Since we prefer to use only voltage as a carrier, and these elements use both voltage and current, we need to convert one quantity to another. The role of the resistor in RC circuits is to make these conversions.

Implementations

RC integrating circuit

The role of the resistor in an RC integrating circuit is to convert the input voltage into current; so it acts here as a voltage-to-current converter. The role of the capacitor is to integrate this current into output voltage; so it is a current-to-voltage integrator (in short, a current integrator). The two converters are cascaded thus forming a voltage-to-voltage integrator (in short, a voltage integrator).

RC differentiating circuit

The role of the capacitor in an RC differentiating circuit is to differentiate the input voltage into current; so it is a voltage-to-current differentiator. The role of the resistor is to convert this current into voltage; so it acts here as a current-to-voltage converter. The two converters are cascaded thus forming a voltage-to-voltage differentiator.

Building an RC integrator

As an example, let's see how this idea is developed in the circuit of an RC integrator.

C integrating circuit

A humble capacitor C supplied by a 1 mA constant current source Iin is a perfect integrator.

^{simulate this circuit – Schematic created using CircuitLab}

The (output) voltage across the capacitor changes linearly through time.

RC integrating circuit

But we prefer to work with voltages rather than currents; that is why we want to apply an input voltage Vin and not an input current to the circuit. So we connect a resistor R in series to Vin to convert the voltage into current.

^{simulate this circuit}

The problem, however, is that the voltage across the capacitor is subtracted from the input voltage. As a result, the current decreases and the curve begins to decrease in steepness. Thus the famous exponent is obtained.

Emulated R"C" circuit

The processes in an RC circuit take place over time, but to understand the main idea it is enough to consider only one moment (e.g. the 7th ms in the graph above), the rest are redundant. For this purpose, we can replace the charged capacitor with a voltage source ("rechargeable battery") of the same voltage (5 V). Thus we can investigate the circuit by ordinary meters and the attractive DC Live Simulation.

^{simulate this circuit}

Manually-compensated R"C" circuit

To fully reveal the role of the resistor, I will continue my story by converting this passive RC circuit into an active one. We can do so by reasoning as follows.

The charged capacitor represents an opposite 5 V positive voltage source with the figurative name C. Its voltage is something harmful that reduces the input voltage and current. Then can't we compensate it by connecting in series a negative source -C with the same voltage of -5 V? We can set it exactly equal to Vc by connecting a null indicator NI to the midpoint between the resistor and the "capacitor", and zeroing it. The name of this technique is negative feedback.

^{simulate this circuit}

As a result, the harmful voltage will be destroyed and the resistor will work at ideal load conditions (zero resistance and voltage); it will "feel" grounded. Another advantage is that the load is powered by the negative "copy" voltage and not by the "original" capacitor voltage. So it will not draw current from the input source.

Automatically-compensated R"C" circuit

Of course, we can simply make a voltage follower do this routine. For this purpose, we can use a VCVS (a conceptual amplifier with K = 1) from the CircuitLab library. We can give it the figurative name "negative capacitor" (because, unlike the "positive" capacitor, it adds its voltage to the input voltage).

^{simulate this circuit}

NFB-compensated R"C" circuit

But we prefer to make a very high-gain (100 k) amplifier do it precisely like us in Schematic 4.1 above (according to the negative feedback principle).

^{simulate this circuit}

NFB-compensated RC circuit

Now that we have grasped the brilliant idea, we can put the actual capacitor in the conceptual circuit above and explore the circuit over time.

^{simulate this circuit}

We see that when the input voltage jumps up to 10 V, the voltage across the capacitor starts to change linearly in the positive direction, and its "copy" in the negative direction. The midpoint voltage is always zero (virtual ground).

Op-amp inverting integrator

Now it only remains to replace the conceptual op-amp with a real one, and we get the famous circuit of an op-amp inverting integrator.

^{simulate this circuit}

We see that the graphs are the same as above.

Conclusions

RC integrator

• In an RC integrating circuit, the resistor acts as an imperfect voltage-to-current converter.

• In an op-amp inverting integrator, the resistor acts as a perfect voltage-to-current converter.

RC differentiator

• In an RC differentiating circuit, the resistor acts as an imperfect current-to-voltage converter.

• In an op-amp inverting differentiator, the resistor acts as a perfect current-to-voltage converter.

More circuits

This "story about a resistor" can be written also for many other circuits:

• RL integrator
• RL differentiator
• voltage divider
• voltage-to-current converter
• current-to-voltage converter
• inverting amplifier
• log converter
• antilog converter, etc.

Answer 3

If you connect a source directly to a capacitor you will have all voltage of the source across the capacitor right at the moment the switch is closed resulting in direc delta function. It's application is in signal processing and quantum mechanics.

By using the resistance in series with the capacitor, the capacitor will be charged relatively slow and it will take some time depending upon the value of R and C to charge the capacitor. That circuit is mainly used as a trigger circuit,filter, integration, differentiation.