# Duty cycle and electronic circuits

A function generator with an internal resistance of $$50\Omega$$ is set to produce a 1kHz square wave with $$20v_{p-p}$$ The duty cycle is 80%. The signal is run through a $$50\Omega$$ resistor.

I need to determine the peak power that is output.

Now, $$\frac{\Delta t}{T}= 0.8$$ so $$\Delta t=0.8T=0.8\left ( \frac{1}{f} \right )=0.0008s$$

But this is as far as I can go.

A useful hint is appreciated. Thanks in advance.

• Is the 50 ohm resistor connected to ground? Is the 20Vp-p into an open circuit? – Andy aka Apr 14 '16 at 8:39
• The peak power doesn't depend on the duty cycle. It does depend on the dc offset, which you haven't specified. – The Photon Apr 14 '16 at 15:01