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The graph shows voltage at several points in a circuit. The points are shown on the x-axis. The y-axis shows the voltage, which is 0 from origin to point a and rises linearly to E from a to b and then drops linearly to E minus I r from b to c. The voltage is constant from c to d and then drops linearly to 0 from d to e.
A graph of the voltage through the circuit of a battery and a load resistance. The electric potential increases the emf of the battery due to the chemical reactions doing work on the charges. There is a decrease in the electric potential in the battery due to the internal resistance. The potential decreases due to the internal resistance ( I r ) , making the terminal voltage of the battery equal to ( ε I r ) . The voltage then decreases by ( IR ). The current is equal to I = ε r + R .

The current through the load resistor is I = ε r + R . We see from this expression that the smaller the internal resistance r , the greater the current the voltage source supplies to its load R . As batteries are depleted, r increases. If r becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates.

Analyzing a circuit with a battery and a load

A given battery has a 12.00-V emf and an internal resistance of 0.100 Ω . (a) Calculate its terminal voltage when connected to a 10.00 - Ω load. (b) What is the terminal voltage when connected to a 0.500 - Ω load? (c) What power does the 0.500 - Ω load dissipate? (d) If the internal resistance grows to 0.500 Ω , find the current, terminal voltage, and power dissipated by a 0.500 - Ω load.


The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated by using the equation V terminal = ε I r . Once current is found, we can also find the power dissipated by the resistor.


  1. Entering the given values for the emf, load resistance, and internal resistance into the expression above yields
    I = ε R + r = 12.00 V 10.10 Ω = 1.188 A .

    Enter the known values into the equation V terminal = ε I r to get the terminal voltage:
    V terminal = ε I r = 12.00 V ( 1.188 A ) ( 0.100 Ω ) = 11.90 V .

    The terminal voltage here is only slightly lower than the emf, implying that the current drawn by this light load is not significant.
  2. Similarly, with R load = 0.500 Ω , the current is
    I = ε R + r = 12.00 V 0.600 Ω = 20.00 A .

    The terminal voltage is now
    V terminal = ε I r = 12.00 V ( 20.00 A ) ( 0.100 Ω ) = 10.00 V .

    The terminal voltage exhibits a more significant reduction compared with emf, implying 0.500 Ω is a heavy load for this battery. A “heavy load” signifies a larger draw of current from the source but not a larger resistance.
  3. The power dissipated by the 0.500 - Ω load can be found using the formula P = I 2 R . Entering the known values gives
    P = I 2 R = ( 20.0 A ) 2 ( 0.500 Ω ) = 2.00 × 10 2 W.

    Note that this power can also be obtained using the expression V 2 R or I V , where V is the terminal voltage (10.0 V in this case).
  4. 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 = ε R + r = 12.00 V 1.00 Ω = 12.00 A .

    Now the terminal voltage is
    V terminal = ε I r = 12.00 V ( 12.00 A ) ( 0.500 Ω ) = 6.00 V ,

    and the power dissipated by the load is
    P = I 2 R = ( 12.00 A ) 2 ( 0.500 Ω ) = 72.00 W.

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

Questions & Answers

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
Emmanuel Reply
sorry I dont know
thank you
What is the pressure?
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
This is at 1 atm
If there is change in pressure u can refer to the steam table ....
instrument for measuring highest temperature of a body is?
brian Reply
how does beryllium decay occur
Sandy Reply
state the first law of thermodynamics
Kansiime Reply
Its state that "energy can neither be created nor destroyed but can be transformed from one form to another. "
what about the other laws can anyone here help with it please
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.
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.
what is distance.?
Ali Reply
what is physics?
Physics is a scientific phenomenon that deals with matter and its properties
physics is the study of nature and science
Physics is branch of science which deals with the study of matters in relation with energy.
What is differential form of Gauss's law?
Rohit Reply
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
Abu Reply
i have to say. who cares. lol. why know that t all
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
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
is the time quantized ? how ?
What do you meanby the statement,"Is the time quantized"
Can you give an explanation.
there are some comment on the time -quantized..
time is integer of the planck time, discrete..
planck time is travel in planck lenght of light..
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
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
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Amount of heat added through q=ncv^delta t
Change in internal energy through q=Q-w
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
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.
Practice Key Terms 5

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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