Problem: A circuit has a source that creates a constant 22 Volt potential difference across series resistances of .01000 Ohms and 1.3 Ohms. What is the current through the source, and how much power is dissipated in the first, and in the second, resistor?
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Solution: Since the current must go through both resistors, it will experience the total resistance .01000 ohm + 1.3ohm = 1.31 ohm. The current must therefore be ( 22 Volts)/( 1.31 Ohms) = 16.79 Amps = 16.79 J/s. The voltage change associated with 16.79 Amps through .01000 ohm is ( 16.79 Amps)( .01000 ohm) = .1679 Volts = .1679 J/C; through the 1.3 ohm resistor the change is ( 16.79 Amps)( 1.3 ohm) = 21.82 Volts = 21.82 J/C. The power dissipated as the 16.79 C/s runs through the first resistor is ( 16.79 C/s)( .1679 J/C) = 2.8 J/s = 2.8 watts, while that through the second resistor is ( 16.79 C/s)( 21.82 J/C) = 370 J/s = 370 watts,
Generalized Response:
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Figure description: The figure below charts the relationships among voltage, resistance, current and power. The 'blue' triangle shows how voltage, current and resistance are related. (Greater voltage or less resistance implies greater current, reflecting I = V / R; resistance is the ratio of voltage to current, reflecting R = V / I; a current I through a resistance R requires a greater voltage drop pfor a greater current and for a greater resistance, reflecting V = I * R).
The 'green' triangle shows how power is the product of current and voltage (current is measured in C / s, voltage in J / C, so the product of current and voltage is the number of J / s, or watts, of power).
The relationship P = I * V can be combined with either I = V / R or with V = I * R to yield either P = I * (I * R) = I^2 * R or P = (V / R ) * V = V^2 / R.
