Unit 7 Problem Solutions

CC2404 Applied Physics and Instrumentation in Health Care Problems Unit 7 (2009) 1. Explain why in a dry weather we may get an electric shock when we touch a metal door. We get some charges on our hand, as metal is conductor, free electrons can flow (current flow) in or out of our hand thus we feel a shock. 2. How can you charge a single metal sphere positively with the help of a negatively charged rubber rod?

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Rubber

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3. (a) What is the energy stored in the 10.0μF capacitor of a heart defibrillator charges to 9,000 volts? (b) Find the amount of stored charges. (a) U=(1/2)CV2=(1/2)x10x10-6x90002=405(J) (b) Q=CV=10x10-6x9000=0.09(C) 4. What current flows through the bulb of a 3.00 V flashlight when its hot resistance is 3.60 Ω ? (0.83 Α) 5. How many volts are supplied to operate an indicator light on a VCR that has a resistance of 140 Ω , given 25.0 mA passes through it? (3.5V) 6. A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path? (30W, 7.5 MΩ) 7. (a) What is the resistance of ten 275Ω r esistors connected in series? (b) In parallel? (2.75 kΩ , 27.5Ω ). a) R=10R1=10*275 b)R=10/R1=10/275 8. (a) What is the resistance of a 100 Ω , a 2.50 k Ω , and a 4.00 Ω resistor connected in series? (b) In parallel? (2604Ω, 3.8Ω) a ) R=100+2500+4 1 1 1 1 = + + = 0.26 , R=3.84Ω b) R 100 2500 4

9. (a) Given a 48.0 V battery, and 24.0 and 96.0 Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel. (a) Rs = R1 + R2 = 24 Ω + 96 Ω = 120 Ω , so that . I =

V 48V = = 0.4 A . Thus, Rs 120 Ω

P1 = I 2 R2 = (0.4 A) 2 (24 Ω) = 3.84W and P2 = I 2 R2 = (0.4 A) 2 (96 Ω) = 15 .4W

(b) I 1 =

V 48V V 48V = = 0.5 A , = = 2 A; I 2 = R2 96 Ω R1 24 Ω

thus: P1 =

V2 ( 48V ) 2 V 2 (48V ) 2 = = 96W , P2 = = = 24W R1 24 Ω R2 96 Ω

10. Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the electron and hydrogen ion to be 9.11x 10-31kg and 1.67 x 10-27 kg. (Notice: ions increase their speed (or kinetic energy) because there is a change in their potential energy) ∆ EPE=qV. 1 2

For electron meVe2 = qV , and for hydrogen

1 m H V H2 = qV 2 1/ 2

me v e2  1.67 ×10 −27 kg  ve mH = 1   = = so that , or −31 vH me m H v H2  9.11 ×10 kg 

= 42 .8