1. An electrosurgical unit produces an open-circuit voltage of 2,500 V (p-p) acr

Question

1. An electrosurgical unit produces an open-circuit voltage of 2,500 V (p-p) across an electrode in the cut mode with a duty cycle of 100%. Compute the power delivered to the load of 250 when the internal resistance of the ESU unit is also 250 . Answer=Power delivered to the load=(Vp-p/2)2/Rint+Rload=(12502/250=6250Watts
2. Follow the same instructions as in the previous problem, except the mode is a blend with a duty cycle of 50%. Compute the power delivered to the load.
1. An electrosurgical unit produces an open-circuit voltage of 2,500 V (p-p) across an electrode in the cut mode with a duty cycle of 100%. Compute the power delivered to the load of 250 when the internal resistance of the ESU unit is also 250 . Answer=Power delivered to the load=(Vp-p/2)2/Rint+Rload=(12502/250=6250Watts
2. Follow the same instructions as in the previous problem, except the mode is a blend with a duty cycle of 50%. Compute the power delivered to the load.
Answer=Power delivered to the load=(Vp-p/2)2/Rint+Rload=(12502/250=6250Watts
2. Follow the same instructions as in the previous problem, except the mode is a blend with a duty cycle of 50%. Compute the power delivered to the load. Answer=Power delivered to the load=(Vp-p/2)2/Rint+Rload=(12502/250=6250Watts
2. Follow the same instructions as in the previous problem, except the mode is a blend with a duty cycle of 50%. Compute the power delivered to the load.

Explanation / Answer

V(p-p)=2500V

Total resistance=500 ohm

So,current I(p-p)=2500/500 A=5 A

So the average current is ,I(avg)=5/sqrt(2)

So the power to the load is P=I^2*R=25/2*250 W=3125 W

Now when the duty cycle is 50%

The average current is, I(avg)=5/(2*sqrt(2)) A

Power is ,P=I^2*R=25/8*250 W=781.25 W

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