Set 53 Problem number 3

Problem

The parallel combination of resistances 36 Ohms and 47 Ohms is maintained at a constant 3 Volt potential difference. What is the current through this circuit, and how much power is dissipated in the first, and in the second, resistor?

Solution

The parallel resistors both have a direct connection to the source. They each therefore experience a voltage drop of 3 Volts.

• This results in a current of 3 Volts / ( 36 ohm) = .08333 Amps = .08333 C/s through the first resistor and of 3 Volts / ( 47 ohm) = .06382 Amps = .06382 C/s through the second.
• These total .1471 Amps, which is the current through the source.

The power dissipated through the first resistor results from the .08333 Coulombs passing through it per second, and the 3 Volt = 3 Joule/Coulomb potential difference across it.

• The result is that ( .08333 C/s) ( 3 J/C) = .2499 J/s = .2499 Watts are dissipated.

Similarly, the second resistor runs through .06382 Coulombs per second, with 3 Joules per Coulomb, resulting in dissipation of .1914 Joules per second, or .1914 Watts.

Generalized Solution

A voltage V across a parallel combination of resistances R1 and R2 will create a potential difference V over both resistances, since both are directly connected to the source. 

• This will result in a current of I1 = V / R1 across the first and a current of I2 = V / R2 across the second.
• The total current will thus be I = I1 + I2 = V / R1 + V / R2 = V * (1 / R1 + 1 / R2).
• It follows that the reciprocal 1 / R of the total resistance is  (1 / R1 + 1 / R2).
• The power across the resistors will be V * I1 = V * V / R1 = V^2 / R1 and V * I1 = V * V / R2 = V^2 / R2.

The power across the circuit will be V * I = V * V * (1 / R1 + 1 / R2) = V^2 * (1 / R1 + 1 / R2).