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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.

Strategy

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.

Solution

  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

if 6.0×10^13 electrons are placed on a metal sphere of charge 9.0micro Coulombs, what is the net charge on the sphere
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18.51micro Coulombs
ASHOK
Is it possible to find the magnetic field of a circular loop at the centre by using ampere's law?
Rb Reply
Is it possible to find the magnetic field of a circular loop at it's centre?
Rb Reply
yes
Brother
The density of a gas of relative molecular mass 28 at a certain temperature is 0.90 K kgmcube.The root mean square speed of the gas molecules at that temperature is 602ms.Assuming that the rate of diffusion of a gas in inversely proportional to the square root of its density,calculate the density of
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A hot liquid at 80degree Celsius is added to 600g of the same liquid originally at 10 degree Celsius. when the mixture reaches 30 degree Celsius, what will be the total mass of the liquid?
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what is electrostatics
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Study of charges which are at rest
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Explain Kinematics
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Using the same two positive charges, the left positive charge is increased so that its charge is 4 times LARGER than the charge on the right. Using your notes on Coulomb's law and changes to the charge, once the charge is increased, what is the new force of repulsion between the two positive charges?
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A mass 'm' is attached to a spring oscillates every 5 second. If the mass is increased by a 5 kg, the period increases by 3 second. Find its initial mass 'm'
Md Reply
a hot water tank containing 50,000g of water is heated by an electric immersion heater rated at 3kilowatt,240volt, calculate the current
Samuel Reply
what is charge
Aamir Reply
product of current and time
Jaffar
Why always amber gain electrons and fur loose electrons? Why the opposite doesn't happen?
Mohammed Reply
A closely wound search coil has an area of 4cm^2,1000 turns and a resistance of 40ohm. It is connected to a ballistic galvanometer whose resistance is 24 ohm. When coil is rotated from a position parallel to uniform magnetic field to one perpendicular to field,the galvanometer indicates a charge
Palak Reply
Using Kirchhoff's rules, when choosing your loops, can you choose a loop that doesn't have a voltage?
Michael Reply
how was the check your understand 12.7 solved?
Bysteria Reply
Who is ISSAAC NEWTON
LOAK Reply
he's the father of 3 newton law
Hawi
he is Chris Issaac's father :)
Ethem
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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