Question

A sinusoidal voltage source, v₁ generating the sinusoid, v₁(t)=V p cos(2π1000t+π/5) is in series with a second AC voltage source, v₂, generating a 100 Hz,50% duty cycle square wave. The amplitude of the sinusoid is, 0.5 V. The square wave has, VH =9.6 V and V L =−7.2 V. The combined voltage, v(t)=v ₁(t) + v₂(t) appearing across the series combination is applied to a 100ohm resistor. What is the DC value of the voltage across the resistor?

Answer 1

The DC value of the voltage across the resistor is 1.2 V, which is calculated by averaging the high and low values of the square wave component since the sinusoidal component has an average of 0 V.

Step-by-step explanation:

The DC value of the voltage across the resistor can be calculated by finding the average value of the combined voltage over one complete cycle of the square wave. Since the sinusoidal voltage source v1(t) has an amplitude of 0.5 V, it will have an average or DC value of 0 V over one cycle because it is a symmetrical waveform centered on 0 V. For the square wave generated by v2, which has a 50% duty cycle, the average value is the midpoint between the high value VH = 9.6 V and the low value VL = -7.2 V. This gives us an average DC value of (9.6 V + (-7.2 V))/2 = 1.2 V. Thus, the DC voltage across the 100 Ohm resistor is 1.2 V.