# 10.1 Electromotive force  (Page 6/11)

 Page 6 / 11

## Summary

• All voltage sources have two fundamental parts: a source of electrical energy that has a characteristic electromotive force (emf), and an internal resistance r . The emf is the work done per charge to keep the potential difference of a source constant. The emf is equal to the potential difference across the terminals when no current is flowing. The internal resistance r of a voltage source affects the output voltage when a current flows.
• The voltage output of a device is called its terminal voltage ${V}_{\text{terminal}}$ and is given by ${V}_{\text{terminal}}=\epsilon -Ir$ , where I is the electric current and is positive when flowing away from the positive terminal of the voltage source and r is the internal resistance.

## Conceptual questions

What effect will the internal resistance of a rechargeable battery have on the energy being used to recharge the battery?

Some of the energy being used to recharge the battery will be dissipated as heat by the internal resistance.

A battery with an internal resistance of r and an emf of 10.00 V is connected to a load resistor $R=r$ . As the battery ages, the internal resistance triples. How much is the current through the load resistor reduced?

Show that the power dissipated by the load resistor is maximum when the resistance of the load resistor is equal to the internal resistance of the battery.

$\begin{array}{}\\ \\ P={I}^{2}R={\left(\frac{\epsilon }{r+R}\right)}^{2}R={\epsilon }^{2}R{\left(r+R\right)}^{-2},\phantom{\rule{0.5em}{0ex}}\frac{dP}{dR}={\epsilon }^{2}\left[{\left(r+R\right)}^{-2}-2R{\left(r+R\right)}^{-3}\right]=0,\hfill \\ \left[\frac{\left(r+R\right)-2R}{{\left(r+R\right)}^{3}}\right]=0,\phantom{\rule{0.5em}{0ex}}r=R\hfill \end{array}$

## Problems

A car battery with a 12-V emf and an internal resistance of $0.050\phantom{\rule{0.2em}{0ex}}\text{Ω}$ is being charged with a current of 60 A. Note that in this process, the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted into chemical energy?

The label on a battery-powered radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-V emf, whereas alkaline cells have a 1.58-V emf. The radio has a $3.20\phantom{\rule{0.2em}{0ex}}\text{Ω}$ resistance. (a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio (b) when using nicad cells, each having an internal resistance of $0.0400\phantom{\rule{0.2em}{0ex}}\text{Ω}$ , and (c) when using alkaline cells, each having an internal resistance of $0.200\phantom{\rule{0.2em}{0ex}}\text{Ω}$ . (d) Does this difference seem significant, considering that the radio’s effective resistance is lowered when its volume is turned up?

a.

b. 0.476W; c. 0.691 W; d. As ${R}_{L}$ is lowered, the power difference decreases; therefore, at higher volumes, there is no significant difference.

An automobile starter motor has an equivalent resistance of $0.0500\phantom{\rule{0.2em}{0ex}}\text{Ω}$ and is supplied by a 12.0-V battery with a $0.0100\text{-}\text{Ω}$ internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add $0.0900\phantom{\rule{0.2em}{0ex}}\text{Ω}$ to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)

(a) What is the internal resistance of a voltage source if its terminal potential drops by 2.00 V when the current supplied increases by 5.00 A? (b) Can the emf of the voltage source be found with the information supplied?

a. $0.400\phantom{\rule{0.2em}{0ex}}\text{Ω}$ ; b. No, there is only one independent equation, so only r can be found.

A person with body resistance between his hands of $10.0\phantom{\rule{0.2em}{0ex}}\text{k}\text{Ω}$ accidentally grasps the terminals of a 20.0-kV power supply. (Do NOT do this!) (a) Draw a circuit diagram to represent the situation. (b) If the internal resistance of the power supply is $2000\phantom{\rule{0.2em}{0ex}}\text{Ω}$ , what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be 1.00 mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning.

A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in $\text{°}\text{C}\text{/}\text{min}$ ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of $0.300\phantom{\rule{0.2em}{0ex}}\text{kcal/kg}\phantom{\rule{0.2em}{0ex}}·\text{°}\text{C}$ , assuming no heat escapes?

a. $0.400\phantom{\rule{0.2em}{0ex}}\text{Ω}$ ; b. 40.0 W; c. $0.0956\phantom{\rule{0.2em}{0ex}}\text{°C/min}$

What mass of steam of 100 degree celcius must be mixed with 150g of ice at 0 degree celcius, in a thermally insulated container, to produce liquid water at 50 degree celcius
sorry I dont know
Bamidele
thank you
Emmanuel
What is the pressure?
SHREESH
To convert 0°C ice to 0°c water. Q=M*s=150g*334J/g=50100 J.......... Now 0° water to 50° water... Q=M*s*dt=150g*4.186J/g*50= 31395 J....... Which adds upto 81495 J..... This is amount of heat the steam has to carry. 81495= M *s=M*2230J/g..therefore.....M=36.54g of steam
SHREESH
This is at 1 atm
SHREESH
If there is change in pressure u can refer to the steam table ....
SHREESH
instrument for measuring highest temperature of a body is?
Thermometer
Umar
how does beryllium decay occur
Photon?
Umar
state the first law of thermodynamics
Its state that "energy can neither be created nor destroyed but can be transformed from one form to another. "
Ayodamola
what about the other laws can anyone here help with it please
Sandy
The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.
sahil
The first law is very simple to understand by its equation. The law states that "total energy in thermodynamic sytem is always constant" i.e d¶=du+dw where d¶=total heat du=internal energy dw=workdone... PLEASE REFER TO THE BOOKS FOR MORE UNDERSTANDING OF THE CONCEPT.
Elia
what is distance.?
what is physics?
Ali
Physics is a scientific phenomenon that deals with matter and its properties
Ayodamola
physics is the study of nature and science
John
Chater1to7
min
Physics is branch of science which deals with the study of matters in relation with energy.
Elia
What is differential form of Gauss's law?
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
kya ye b.sc ka hai agar haa to konsa part
what is charge quantization
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
determine absolute zero
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
what is copper loss
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry