Electrosurgical Units Electrosurgical units heat tissue to achieve desired surgi

Question

Electrosurgical Units Electrosurgical units heat tissue to achieve desired surgical effects, namely cutting, desiccation or coagulation. The change in tissue temperature is given by the following formula: J: current density, t: activation time, p: tissue density, o: electrical conductivity, c: specific heat capacity. (a) Describe strategies a surgeon may use to achieve the desired surgical effect (cutting, desiccation or coagulation), relating the strategy with the factors in the formula (J, t, etc.) (b) Burns are a risk to the patient when using an ESU. Describe the instrumentation used to mitigate this risk (your answer should discuss this in the context of current density).

Explanation / Answer

a. given

dT = J^2t/sigma*rho*C

J is current density

t is activation time

sigma is conmductivity

rho is tissue density

c is specific heat capacity

dT is change in temeprature

for desiccation,the tissue needs to get dry (lose water content)

for coagulation, the protien structure in the tissue shall change, causing permanent changes

for cutting, very high tissue temperaure is required

hence

for dessication, we shall have lowest values of J spread over an area of tissue uniform;y

for coagulation, we shall have higher values of J spread over lesser area ( for more temperature rise)

for cutting, we need to concentrate high values of J on low areas to have high dT

b. While the electrosurgical unit (ESU) is commonly used during surgical procedures, there are precautions that must be addressed to prevent injuries. The ECRI Institute has published two safety reports addressing patient injuries related to poor return electrode site preparation. Factors that can contribute to electrosurgical burns include poor electrode placement, inadequate site preparation, and defective return electrodes

Defibrilation

L = 0.025 H

R = 75 ohm

C = ?

hence

Lq" + Rq' + q/C = 0

now for critical damping

R^2 = 4L/C

C = 4L/R^2 = 17.77 uF

peak current = Io

now

E = 200 J = CV^2/2

V = 4743.416490 V

Io = V/R = 63.24555320 A

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