Problem: A circuit has a source that creates a constant 3.681 Volt potential difference across parallel resistances of 1.178 Ohms and 2.848 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: The parallel resistors both have a direct connection to the source. They each therefore experience a voltage drop of 3.681 Volts. This results in a current of 3.681 Volts / ( 1.178 ohm) = 3.124 Amps = 3.124 C/s through the first resistor and of 3.681 Volts / ( 2.848 ohm) = 1.292 Amps = 1.292 C/s through the second. These total 4.416 Amps, the current through the source. The power dissipated through the first results from the 3.124 Coulombs passing through it per second, and the 3.681 Volt = 3.681 Joule/Coulomb potential difference across it. The result is that ( 3.124 C/s) ( 3.681 J/C) = 11.49 J/s = 11.49 Watts are dissipated. Similarly, the second resistor runs through 1.292 Coulombs per second, with 3.681 Joules per Coulomb, resulting in dissipation of 4.755 Joules per second, or 4.755 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.
