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Question. No: 1

 

a)      Is it true to say that, current decreases when passing through a resistor and then increases again upon exiting?

b)      Is it true or wrong to say that, Magnetic fields exert a force on all moving electrically charged particles? Either yes or no explain it.                                                                                                                                                                            

 

Question. No: 2

The defibrillator capacitor discharges in 15 ms (milli second) as shown in below figure. How much current does it send through the heart?                                                   

Question. No: 3

 

A power plant provides =500 MW of power through transmission lines at a potential difference of 500 kV. What current flows through the lines? For a city 360 km away from the power plant the total resistance in the transmission lines is 70 ohms. What fraction of the generated power is dissipated as “line loss”? If the generated power is transmitted at a potential difference of 185 kV instead, what fraction is dissipated??                                                                            s

 

 

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Question:01
Part-01
Resistors have two primary functions within a circuit:
Reduce current flow
Lower voltage
Resistors are usually made of metal or carbon wire. They are designed to keep a constant value when in use and produce heat instead of light when used inside an electrical component. As the circuit works, resistors dissipate heat, which is not used. On the other hand, the metal portion of a light bulb is a type of resistor that does emit light.
Part-02
Yes Magnetic fields exert a force on all moving electrically charged particles.
Question:02
The defibrillator capacitor discharges in 15 ms (milli second) as shown in bellow figure. How much current does it send through the heart?
Solution:
C=88μF = 88×10-6F
ΔV=3.5KV=3.5×103V
Time=t=15 ms = 15×10-3s
C=q/ΔV=q= CΔV
(88×10-6)(3.5×103 V)
I=Δq/Δt =−3.08×107/15×10-3s
=205.34A
Question:03
P = V*I ------> I = P/V = 500e6/500e3 = 1000A assumed through ONE line.
I²*R = 1e6*70 = 70e6
70e6/500e6 = 7/50 = 14%
I²R/P = (P/V)²R/P = P*R/V² = 500e6*70/185kv² = 1400/1369 = 1.02 = 102%
So slightly more power is lost than generated

PHY101 assignment solution

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