Homework Sourse1 An electrosurgical unit produces an opencircuit voltage of
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.
Solution
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
