Unit 8 Problems Solution

CC2404 Applied Physics and Instrumentation in Health Care Problems Unit 8 (2008)

1. Which ion rushes into a resting cell when depolarization occurs? (Na ions) 2. What is the normal resting potential across of a human nerve cell? (-70 to -90 mV) 3. What is the approximate amplitude of a ECG pattern recorded from the skin? ( 1 mV) 4.

For a typical ECG waveform, what are the meanings of P- , QRS- and T- waves? The P-wave: occurring during depolarization of the atria, causing them to contract.

The QRS-wave: corresponding to the depolarization and contraction of the ventricles. The T-wave: resulting from re-polarization and relaxation of the ventricles. During surgery, a current as small as 20.0 µ A applied directly to the heart may cause

5.

ventricular fibrillation. Do you think that special electrical safety precautions are needed?

It would not be necessary to take extra precautions regarding the power coming from the wall. However it is possible to generate voltages of approximately this value from the static charge built up on gloves, for instance, so some precautions are necessary.

6. The outer space of a neuron’ s axon membrane (κ =5 thickness d= 1.2x10-8m ) is charged positively, and the inner portion is charged negatively. Assuming that an axon can be treated like a parallel plate capacitor with a plate area of 6x10-6 m2, what is its capacitance ? (22.1 nF)

C = ε oκ

7.

A 6 ×10 −6 = 8.85 ×10 −12 × 5 × d 1.2 ×10 −8

(a) During surgery, a current as small as 20.0 µ A applied directly to the heart may cause

ventricular fibrillation. If the resistance of the exposed heart is 300 Ω , what is the smallest voltage that poses this danger? (b) For a time of 10 s, how much charge passed through the heart? (c) What is the energy dissipated during this 10 s? (a)V=IR=20 x 10-6A x 300 Ω =0.006 V=6 mV (b) Q=It=20 x 10-6 x 10=0.0002C=2 x 10-4C (c) E=IVt=QV=2 x 10-4 x 6x10-3=1.2 x 10-6J

8. a) What is the peak current through a 500 W room heater that operates on 120 V AC power? (b)What is the peak power? (c) For the same power consumption, if it operates on 220 V AC line, what is the peak current. (5.89A, 1000W, 3.2A) Pave 500 500 = a) I rms = , I o = 2 I rms = 2 × Vrms 120 120

b) Po=2*Pave=2*500=1000W c) I rms =

Pave 500 500 = , I = 2 ⋅ I rms = 2 × Vrms 220 o 220

9. (a) What is the resistance of a 220 V ac short circuit that generates a peak power of 96.8 kW? (b) What would be the average power be if the voltage was 120 V ac? (a) Pave =

Po V2 = 48400 = rms 2 R

P V  (b) 2 =  2  P1  V1 

2

V P2 = P1  2  V1

R=

2 Vrms 220 2 = = 1.00 Ω Pave 48400

2

  = 14 .4kW 

10. What is the resistance of a copper wire 0.5 mm in diameter and 20 m long? The resistivity of copper is 1.7 x 10-8 Ω·m. The wire’s cross-sectional area is π r2, where r = 0.25 mm = 2.5 x 10-4. Hence R =ρ

L (1.7 ×10 −8 Ω ⋅ m)( 20 m) = = 1.73 Ω A (π )( 2.5 ×10 −4 m) 2

11. Estimate the current and power dissipation on a human being under a voltage of 1 V and 1000 V respectively. We can model the human body into a cylindrical geometry of 30 cm in diameter and 1 m in height. We assume a resistivity of 500 Ω -cm for the tissue. L 100 R = ρ ⋅ = 500 ⋅ = 70.7Ω A π 15 2 For 1 V: I=V/R=14 mA, P=IV=0.014x1=14mW For 1000V: I=V/R=14 A, P=IV=14x1000=14kW If V increases 1000 times, power does 1000,000 times.