(decay time). Since the waveform does not level off at a constant

value, the duration time is zero. The triangular waveform thus is

different from the square wave in which the voltage rises quickly and

then levels off. Likewise, the response of RC and RL circuits to a

triangular wave is different from their response to a square wave.

b. Response of an RC circuit. The response of an RC circuit to a

triangular wave input is illustrated in Figure 30. Figure 30C shows

the voltage waveforms enlarged several times for easier study. When

a square wave is applied, the capacitor charges gradually towards the

value at which the square wave levels off; with the application of a

triangular wave, the voltage continues to increase as the capacitor

charges toward it, and the capacitor must continuously charge toward

a new, higher value. Initially, when the voltage E is applied to the

circuit, the current rises with the voltage because the capacitor is

uncharged. The current charges the capacitor to a voltage that

opposes the applied voltage. This action should decrease the circuit

current, but since the input voltage is continuously increasing, the

opposing voltage across the capacitor is more than overcome by the

applied voltage, and the current actually increases, but at a slower

rate than it did initially. As the current continues to increase,

the capacitor becomes charged to a higher voltage and the rate of

current increase is reduced further until, finally, the current is

high enough to raise the capacitor voltage at the same rate as the

increase in the input waveform. At this point, the current becomes

constant, and the capacitor voltage slope is essentially the same as

the input voltage slope.

(1) The current curve is the same as the curve ER. At the end

of 1 time constant, the resistor voltage is equal to approximately

63.2 percent of the voltage applied at that time, and the current is

equal to 63.2 percent of the voltage applied at that time divided by

the value of the resistance in the circuit. After 7 time constants,

ER will level off at a value equal to the voltage applied at the end

of the firsttime constant, and the current will be equal to this

voltage divided by the resistance. The capacitor continues to charge

at the same rate as the rise in input voltage, until the rise of

decreasing during the second half of the cycle, the charging rate of

the capacitor decreases, until the applied voltage drops to the same

value as the capacitor voltage (Figure 31B). At this point, the

current in the circuit is zero. As the applied voltage continues to

decrease, it falls below the capacitor voltage, and the capacitor

starts to discharge.

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