# Power and measuring itPage 3

#### WATCH ALL SLIDES

If a motor lifts the water

from a depth ‘h1’ and then

raises to a height ‘h2’ then

the power of the motor is

given by

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RGUIIIT BASAR

Slide 17 ## Power of heart

Power of heart = work/time

= (F.s)/t

= (PxA.s)/t

= (PxV )/t

= (hdgV)/t

(P = pressure,A = area of vessel, V = volume of vessel and

s = length of the vessel)

Thus power of heart = P(V/t) = (hdgV)/t

= Pressure x volume of blood pumped per second

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RGUIIIT BASAR

Slide 18 ## Heart problem

The heart of a man pumps 4 litres of blood per minute at a pressure of 130 m.m. of Hg. If the density of the blood is 13.6 gm/c.c. calculate the power of the heart.

Solution:-

Power of heart = (hdgV)/t

= 1.155watt.

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RGUIIIT BASAR

Slide 19 ## Power of lungs

Power of lungs =

(mass of air blown per second ) x (velocity)2

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RGUIIIT BASAR

Slide 20 ## Work in terms of power

The work done by from time t1 to time t2 is given by

Where P = F.v

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RGUIIIT BASAR

W

Slide 21 ## Power

If force F acted on a body of mass ‘m’ which is at rest , then the power produced in that body in time ‘t’ is given by

P = F.v

= F(F/m)t

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RGUIIIT BASAR

V = u + at

= 0 + at

= at

= (F/m)t

P

P = Fv

= mav

= ma(at)

= ma2t

Slide 22 A box of mass ‘m’ moved along a straight line by a machine delivering constant power(P).Then the distance moved by the body in terms of m, P & t is given by

Solution:- P=Fv = mav

P= m(dv/dt)v

vdv = (P/m)dt ,by integrating we get

dx/dt

On integrating we get,

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RGUIIIT BASAR

Slide 23 ## Electrical power

Electrical power

Instantaneous electrical power

The instantaneous electrical power P delivered to a component is given by

P(t) = V(t).I(t)

where P(t) is the instantaneous power, measured in watts (joules per second)

V(t) is the potential difference (or voltage drop) across the component, measured in volts

I(t) is the current through it, measured in amperes

If the component is a resistor, then:

P = V.I = I2 .R

where R is the resistance, measured in ohms.

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