A full-wave rectified voltage of \(18 \mathrm{~V}\) peak is connected to a \(400-\mu \mathrm{F}\) filter capacitor. What are the ripple and dc voltages across the capacitor at a load of \(100 \mathrm{~mA}\) ?

Short Answer

Expert verified
The ripple voltage is approximately 15V and the DC voltage is 18V.

Step by step solution

01

Calculate the DC Voltage

The DC voltage (also known as the peak voltage) is provided in the problem statement as \(18V\). So, the DC voltage across the capacitor, \(V_{dc}\), is thus \(18V\).
02

Calculate frequency in seconds

AC mains supply frequency is 50 or 60 Hz is most countries. Since it is not mentioned, we will use the 60 Hz as it is most common and because the unit of frequency is cycles per second, convert the frequency \(f\) into seconds as: \(f = 1/60 s^{-1}\).
03

Calculate the Ripple Voltage

Now we can calculate the ripple voltage, represented as \(V_r\). The formula is: \(V_r = I/ (f * C) \), where \(I\) is the load current, \(f\) is the frequency and \(C\) is the capacitance. Substituting the given values, we have \(V_r = 100*10^{-3}/(1/60*400*10^{-6})\). After calculating, we find that \(V_r \approx 15V\).

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