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Phys102 Lecture 9 Electric Currents and Resistance

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1 Phys102 Lecture 9 Electric Currents and Resistance
Key Points Ohm’s Law Resistivity Electric Power Alternating Current References 18-1,2,3,4,5,6,7

2 Unit of electric current: the ampere, A:
Electric current is the rate of flow of charge through a conductor: The instantaneous current is given by: Unit of electric current: the ampere, A: 1 A = 1 C/s.

3 Ohm’s Law The ratio of voltage to current is called the resistance:
1.8kW 1.5V RL VL IL

4 Resistivity The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area: The constant ρ, the resistivity, is characteristic of the material. A l

5 Resistivity This table gives the resistivity and temperature coefficients of typical conductors, semiconductors, and insulators.

6 Electric Power Power, as in kinematics, is the energy transformed by a device per unit time: or

7 Electric Power The unit of power is the watt, W.
For ohmic devices, we can make the substitutions:

8 Electric Power Example: Headlights.
Calculate the resistance of a 40-W automobile headlight designed for 12 V. Solution: R = V2/P = 3.6 Ω.

9 Alternating Current Current from a battery flows steadily in one direction (direct current, DC). Current from a power plant varies sinusoidally (alternating current, AC). Figure (a) Direct current. (b) Alternating current.

10 Alternating Current The voltage varies sinusoidally with time:
, , as does the current:

11 Alternating Current Multiplying the current and the voltage gives the power: Figure Power transformed in a resistor in an ac circuit.

12 Alternating Current Usually we are interested in the average power: .

13 Alternating Current The current and voltage both have average values of zero, so we square them, take the average, then take the square root, yielding the root-mean-square (rms) value:

14 Alternating Current Example: Hair dryer.
(a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain? Figure A hair dryer. Most of the current goes through the heating coils, a pure resistance; a small part goes to the motor to turn the fan. Example 25–13. Solution: a. The rms current is the power divided by the rms voltage, or 8.33 A, so the peak current is 11.8 A. Then R = V/I = 14.4 Ω. b. Assuming the resistance would stay the same (probably wrong – the heat would increase it), doubling the voltage would mean doubling the current, and the power (P = IV) would increase by a factor of 4. This would not be good for the hair dryer.


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