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Power ( P ) is exactly equal to current ( I ) multiplied by voltage ( V ) and there is no extra constant of proportionality. The unit of measurement for power is the Watt (abbreviated W).

Interesting fact

It was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the equation:
P = I 2 R
and is properly known as Joule's Law. However, these power equations are so commonly associated with the Ohm's Law equations relating voltage, current, and resistance that they are frequently credited to Ohm.

Investigation : equivalence

Use Ohm's Law to show that:

P = V I

is identical to

P = I 2 R

and

P = V 2 R

Summary

  1. Ohm's Law states that the amount of current through a conductor, at constant temperature, is proportional to the voltage across the resistor. Mathematically we write V = I · R
  2. Conductors that obey Ohm's Law are called ohmic conductors; those that do not are called non-ohmic conductors.
  3. We use Ohm's Law to calculate the resistance of a resistor. R = V I
  4. The equivalent resistance of resistors in series ( R s ) can be calculated as follows: R s = R 1 + R 2 + R 3 + . . . + R n
  5. The equivalent resistance of resistors in parallel ( R p ) can be calculated as follows: 1 R p = 1 R 1 + 1 R 2 + 1 R 3 + . . . + 1 R n
  6. Real batteries have an internal resistance.
  7. Wheatstone bridges can be used to accurately determine the resistance of an unknown resistor.

End of chapter exercise

  1. Calculate the current in the following circuit and then use the current to calculate the voltage drops across each resistor.
  2. Explain why a voltmeter is placed in parallel with a resistor.
  3. Explain why an ammeter is placed in series with a resistor.
  4. [IEB 2001/11 HG1] - Emf
    1. Explain the meaning of each of these two statements:
      1. “The current through the battery is 50 mA.”
      2. “The emf of the battery is 6 V.”
    2. A battery tester measures the current supplied when the battery is connected to a resistor of 100 Ω . If the current is less than 50 mA, the battery is “flat” (it needs to be replaced). Calculate the maximum internal resistance of a 6 V battery that will pass the test.
  5. [IEB 2005/11 HG] The electric circuit of a torch consists of a cell, a switch and a small light bulb, as shown in the diagram below.
    The electric torch is designed to use a D-type cell, but the only cell that is available for use is an AA-type cell. The specifications of these two types of cells are shown in the table below:
    Cell emf Appliance for which it is designed Current drawn from cell when connected to the appliance for which it is designed
    D 1,5 V torch 300 mA
    AA 1,5 V TV remote control 30 mA
    What is likely to happen and why does it happen when the AA-type cell replaces the D-type cell in the electric torch circuit?
    What happens Why it happens
    (a) the bulb is dimmer the AA-type cell has greater internal resistance
    (b) the bulb is dimmer the AA-type cell has less internal resistance
    (c) the brightness of the bulb is the same the AA-type cell has the same internal resistance
    (d) the bulb is brighter the AA-type cell has less internal resistance
  6. [IEB 2005/11 HG1] A battery of emf ε and internal resistance r = 25 Ω is connected to this arrangement of resistors.
    The resistances of voltmeters V 1 and V 2 are so high that they do not affect the current in the circuit.
    1. Explain what is meant by “the emf of a battery”. The power dissipated in the 100 Ω resistor is 0,81 W.
    2. Calculate the current in the 100 Ω resistor.
    3. Calculate the reading on voltmeter V 2 .
    4. Calculate the reading on voltmeter V 1 .
    5. Calculate the emf of the battery.
  7. [SC 2003/11] A kettle is marked 240 V; 1 500 W.
    1. Calculate the resistance of the kettle when operating according to the above specifications.
    2. If the kettle takes 3 minutes to boil some water, calculate the amount of electrical energy transferred to the kettle.
  8. [IEB 2001/11 HG1] - Electric Eels Electric eels have a series of cells from head to tail. When the cells are activated by a nerve impulse, a potential difference is created from head to tail. A healthy electric eel can produce a potential difference of 600 V.
    1. What is meant by “a potential difference of 600 V”?
    2. How much energy is transferred when an electron is moved through a potential difference of 600 V?

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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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