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# Physics Assignment No 3 Solution, Phy101 DueDate 26 Jan

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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Q.2

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

∆V=3.5KV=3.5×103V

Time=t=15 ms = 15×10-3s

C=q/∆V=q= C∆V

(88×10-6)(3.5×103 V)

I=∆q/∆t =−3.08×107/15×10-3s

=205.34A

anyone who solves numerical?? ans confirm krna

Q.2

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

∆V=3.5KV=3.5×103V

Time=t=15 ms = 15×10-3s

C=q/∆V=q= C∆V

(88×10-6)(3.5×103 V)

I=∆q/∆t =−3.08×107/15×10-3s

=205.34A

Question 3 ka answer agar kisi k pas ho to please share kr djye

usma current find karain then

power(loss) which is I^2R accordingly..

Koi toh question 3 ka answer post kreeeeeeeee..

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

Q.2

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

∆V=3.5KV=3.5×103V

Time=t=15 ms = 15×10-3s

C=q/∆V=q= C∆V

(88×10-6)(3.5×103 V)

I=∆q/∆t =−3.08×107/15×10-3s

=205.34A

phy101 assignment solution......!!

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