Half wave rectifier output

In the half wave rectifier circuit with a capacitor filter i'm getting an output of 26 volts DC for 24 volts AC input.....i want to know how this is possible....output is more than the input....i used 1n4007 PN junction diode,1K resistor and 100uF cap....Please clarify my doubt...

24Vac means 24VRMS. The peak value is 24√2= 33.9Vp. The rectifier tends to charge the output filtering capacitor to the peak value. Furthermore, you´ll se some ripple on it. Better not trusting in your DMM to measure that. Please google for RMS and peak value. Also, the output voltage of the transformer will change depending on the load.

Can you please be a bit more clear....i cant understand it...if the capacitor is charged upto the peak value,how is the output exceeding the input???,why cant a DMM be trusted for measuring....please explain...please

The output of the transformer is 24Vac (=24VRMS). Use your DMM in AC mode to measure it. A sine wave of 24VRMS has a peak voltage of 33.9Vpeak. So, after rectification, your filtering capacitor should be charged to 33.9Vdc-Vf, where Vf is the voltage drop due to the rectifier. This is the expected DC voltage at the output with no load. Use your DMM in DC mode to measure it. However, when there is some load two things happen:
1- The output voltage of the transformer will decrease (loss in the coil´s resistance)(not to mention current imbalance in the transformer´s coil).
2- The voltage at the output capacitor will show some ripple (inverse to the load value). So, using a DMM you´ll measure the average output voltage and not the ripple.
Hope this is enough.
Regards

yeah its clear...thanks a lot....so the value i measure as the output is less than the peak input voltage....so there is nothing absurd in it.....am i right???

If it is loaded the measured value seems to be right.

yeah....clear....thanks a lot....hope i gain a lot of knowledge from you and this forum....thank you very mush....