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V = emf Ir = 12.0 V ( 1.188 A ) ( 0.100 Ω ) = 11.9 V.

Discussion for (a)

The terminal voltage here is only slightly lower than the emf, implying that 10 . 0 Ω size 12{"10" "." 0 %OMEGA } {} is a light load for this particular battery.

Solution for (b)

Similarly, with R load = 0 . 500 Ω size 12{R rSub { size 8{"load"} } =0 "." "500"` %OMEGA } {} , the current is

I = emf R load + r = 12 . 0 V 0 . 600 Ω = 20 . 0 A . size 12{I= { {"emf"} over {R rSub { size 8{"load"} } +r} } = { {"12" "." 0" V"} over {0 "." "600 " %OMEGA } } ="20" "." 0" A"} {}

The terminal voltage is now

V = emf Ir = 12.0 V ( 20.0 A ) ( 0.100 Ω ) = 10 . 0 V.

Discussion for (b)

This terminal voltage exhibits a more significant reduction compared with emf, implying 0 . 500 Ω size 12{0 "." "500 " %OMEGA } {} is a heavy load for this battery.

Solution for (c)

The power dissipated by the 0 . 500 - Ω size 12{0 "." "500-" %OMEGA } {} load can be found using the formula P = I 2 R size 12{P=I rSup { size 8{2} } R} {} . Entering the known values gives

P load = I 2 R load = ( 20.0 A ) 2 ( 0 .500 Ω ) = 2.00 × 10 2 W . size 12{P rSub { size 8{"load"} } =I rSup { size 8{2} } R rSub { size 8{"load"} } = \( "400"" A" rSup { size 8{2} } \) \( 0 "." "500" %OMEGA \) ="200"" W"} {}

Discussion for (c)

Note that this power can also be obtained using the expressions V 2 R size 12{ { {V rSup { size 8{2} } } over {R} } } {} or IV size 12{ ital "IV"} {} , where V size 12{V} {} is the terminal voltage (10.0 V in this case).

Solution for (d)

Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding

I = emf R load + r = 12 . 0 V 1 . 00 Ω = 12 . 0 A . size 12{I= { {"emf"} over {R rSub { size 8{"load"} } +r} } = { {"12" "." 0" V"} over {1 "." "00 " %OMEGA } } ="12" "." 0" A"} {}

Now the terminal voltage is

V = emf Ir = 12.0 V ( 12.0 A ) ( 0.500 Ω ) = 6.00 V,

and the power dissipated by the load is

P load = I 2 R load = ( 12.0 A ) 2 ( 0 . 500 Ω ) = 72 . 0 W . size 12{P rSub { size 8{"load"} } =I rSup { size 8{2} } R rSub { size 8{"load"} } = \( "144"" A" rSup { size 8{2} } \) \( 0 "." "500" %OMEGA \) ="72" "." 0" W"} {}

Discussion for (d)

We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load.

Battery testers, such as those in [link] , use small load resistors to intentionally draw current to determine whether the terminal voltage drops below an acceptable level. They really test the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by its low terminal voltage.

The first photograph shows an avionics electronics technician working inside an aircraft carrier, measuring voltage of a battery with a voltmeter probe. The second photograph shows the small black battery tester which has an LED screen that indicates the terminal voltage of four batteries inserted into its case.
These two battery testers measure terminal voltage under a load to determine the condition of a battery. The large device is being used by a U.S. Navy electronics technician to test large batteries aboard the aircraft carrier USS Nimitz and has a small resistance that can dissipate large amounts of power. (credit: U.S. Navy photo by Photographer’s Mate Airman Jason A. Johnston) The small device is used on small batteries and has a digital display to indicate the acceptability of their terminal voltage. (credit: Keith Williamson)

Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to a resistance. This is done routinely in cars and batteries for small electrical appliances and electronic devices, and is represented pictorially in [link] . The voltage output of the battery charger must be greater than the emf of the battery to reverse current through it. This will cause the terminal voltage of the battery to be greater than the emf, since V = emf Ir size 12{V="emf" - ital "Ir"} {} , and I size 12{I} {} is now negative.

The diagram shows a car battery being charged with cables from a battery charger. The current flows from the positive terminal of the charger to the positive terminal of the battery, through the battery and back out the negative terminal of the battery to the negative terminal of the charger.
A car battery charger reverses the normal direction of current through a battery, reversing its chemical reaction and replenishing its chemical potential.

Multiple voltage sources

There are two voltage sources when a battery charger is used. Voltage sources connected in series are relatively simple. When voltage sources are in series, their internal resistances add and their emfs add algebraically. (See [link] .) Series connections of voltage sources are common—for example, in flashlights, toys, and other appliances. Usually, the cells are in series in order to produce a larger total emf.

Questions & Answers

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
kwame Reply
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
Hayne Reply
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir
Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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