Answer to Question #199511 in Electric Circuits for Anuj

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

Explanations & Calculations

a)

• The area of the electrode is

$\qquad\qquad \begin{aligned} \small A&=\small 1.50\,m\times 1.80\,m\\ &=\small 2.7\,m^2 \end{aligned}$

• Then the current density during a first procedure

$\qquad\qquad \begin{aligned} \small j_1&=\small \frac{i}{A}=\frac{10\,mA}{2.7m^2}=\bold{3.7\,mA\,m^{-2}} \end{aligned}$

• That during a second procedure

$\qquad\qquad \begin{aligned} \small j_2&=\small \frac{15}{2.7}=\bold{5.6\,mA\,m^{-2}} \end{aligned}$

b)

• If the resistance of the tissue is R, it remains the same for both cases.
• Then heat generation during the first 5 procedures is

$\qquad\qquad \begin{aligned} \small E_1&=\small i^2Rt\\ &=\small (10\times10^{-3}A)^2\times R\times (5\times 10\times60s)\\ &=\small 0.3R \end{aligned}$

• That during the second 15 procedures is

$\qquad\qquad \begin{aligned} \small E_2&=\small (15\times10^{-3})^2\times R\times (15\times 15\times60s)\\ &=\small 3.04R \end{aligned}$

• Then by comparison we get

$\qquad\qquad \begin{aligned} \small \frac{E_2}{E_1}&=\small \frac{3.04R}{0.3R}=10.1 \end{aligned}$

• It could be seen that the heat generation during the second procedure is approximately 10 times the first procedure.