Answer to Question #199250 in Electric Circuits for Tarik

Question

Electrophoresis is a medical therapy technique used to deliver medicaments into the patients body with help of an electric current. During the procedure, two oppositely charged plates are applied to the body providing flow of the current through the tissue. Suppose that current used in first five procedures was 10 mA (duration of the procedure 10 min) and during the last fifteen procedures 15 mA (duration 20 min). (a) Find current density values for both cases if electrodes area was 𝐴 = 150×180 cm2. (b) Compare heat effects of the current during the first procedure and during the last procedure.

Answer 1

a)

Case 1

$J=\frac{I}{A}=\frac{10*10^{-6}}{1.5*1.8}=0.0000037037 Am^{-2}$

Case 2

$J=\frac{I}{A}=\frac{15*10^{-6}}{1.5*1.8}=0.00000555556 Am^{-2}$

b) If we let R =1$\Omega$

Case 1

$H=I^2Rt=(10*10^{-6})^2*10*60*1=0.00000006 J$

Case 2

$H=I^2Rt=(15*10^{-6})^2*20*60*1=0.00000027 J$

Therefore the heat flux of case 2 is more that that of case 1