Home > Articles

  • Print
  • + Share This
  • 💬 Discuss
This chapter is from the book

Charge and Discharge Currents

There are a few waveforms and relationships that people in our industry should be able to recognize and sketch from memory. Figures 3-10 through 3-12 represent some of these. Figure 3-10 illustrates a simple resistor-capacitor (RC) circuit. When the switch is one position, current flows toward the capacitor. When the switch is in the other position, the capacitor discharges.

03fig10.gifFigure 3-10. Typical charging and discharge voltage and currents for an RC circuit.

03fig11.gifFigure 3-11. Typical charging and discharge voltage and currents for an RL circuit.

03fig12.gifFigure 3-12. Voltage (solid line) leads current through an inductor (a) and voltage lags current through a capacitor (b).

When we first connect the battery, current flows toward the capacitor. In fact, from Ohm's Law, if R = 1.0 the flow will be 1.0V/1.0 Ω = 1.0 amp. Remembering that for capacitors, current leads voltage, and since the capacitor initially has no charge on it, even though the current flow is 1.0 amp, there will be 0.0 volts across the capacitor. Therefore, 1.0 volt must appear across the resistor (1.0 amp through 1.0Ω).

As current flows onto the plate of the capacitor, the voltage across the capacitor must begin to rise. Since the battery voltage does not change, the voltage across the resistor must necessarily decrease (the sum of the voltages across the capacitor and across the resistor must be constant and, in this case, equal to the voltage of the battery, 1.0 volt). Since the value of the resistor does not change, then the current through it must begin to decrease. After some period of time, the current will decline to zero and the voltage across the capacitor will rise to equal that at the battery.

The graph in Figure 3-10 illustrates the relationship. As voltage across the capacitor increases, current decreases. All RC circuits follow a curve such as this, and differ only in terms of the height of the curves or how "stretched out" they are along the horizontal axis. But the shape of the curve is always that of an exponential, and we will define the equation for it in Chapter 4 in the section on time constants.

When we flip the switch to remove the battery from the circuit, the capacitor then discharges through the resistor. Note that the fully charged capacitor (1.0 volt) now causes a current of 1.0 amp (initially) to flow through the resistor in the opposite direction. Thus, the voltage across the resistor is now of the opposite sign. As before, the curves follow a very well-defined pattern. In fact, in this type of situation (a fixed DC source, a simple discharge path around the battery, and time to reach equilibrium) the discharge curves must be identical mirror images of the charging curves.

Figure 3-11 shows the corresponding RL (resistor-inductor) circuit. The analysis is identical. When the battery is first connected, the voltage across the inductor jumps up to 1.0 volt and there is no current through the inductor. Then as current begins to flow through the inductor, the voltage across the inductor begins to decline. At equilibrium, there is no voltage across the inductor, and the current is defined by the battery and the resistor:

058equ01.gif


After the switch is thrown, disconnecting the battery from the circuit, the inductor tries to maintain the same current flow. Since the current doesn't change instantaneously, the voltage across the resistor doesn't change. So the voltage remains at 1.0 volt across the resistor. The left side of the resistor is now the same node as the bottom of the inductor (since the switch has switched), so the voltage across the inductor is now the opposite of what it was when the switch was first connected to the battery.

Again, these curves follow a well-defined exponential relationship, and the discharge curves are mirror images of the charging curves.

When a sine wave current source is applied to a capacitor or an inductor, the voltage across the device also follows a sine wave. For pure capacitors and inductors (i.e., no resistance component at all) the voltage waveform will be shifted (exactly) 90 degrees from the driving current waveform. (Remember, there are 360 degrees in one cycle.) This result can be proven mathematically, but the proof requires mathematical analyses well beyond the scope of this book. So you'll just have to take this fact on faith.

For capacitors, current will lead voltage by 90 degrees (Figure 3-12a) and for inductors it lags voltage by 90 degrees (Figure 3-12b). Conversely, for capacitors, voltage lags current by 90 degrees and for inductors voltage leads current by 90 degrees. If we connect a capacitor and an inductor in series, current must necessarily be the same through both of them (current is the same everywhere along a loop). Therefore, the voltage across the capacitor would lag the voltage across the inductor by 180 degrees (or exactly one-half cycle). Later on we will look at the total voltage across the inductor–capacitor (LC) series pair and note that since their voltages are 180 degrees out of phase, one subtracts from the other. A particularly interesting case is when the voltages across each component are equal, but 180 degrees opposite in phase. Even though there can be voltage across each component, the voltage across the pair will be equal and opposite at all times. Therefore, the combined voltage across the pair will be zero! This surprising result can have very important consequences for certain types of circuits.

  • + Share This
  • 🔖 Save To Your Account

Discussions

comments powered by Disqus