Set 53 Problem number 2

Problem

A circuit has a source that creates a constant 8 Volt potential difference across series resistances of 28 Ohms and 21 Ohms. What is the current through the source, and how much does the voltage change across the first, and across the second, resistor?

Solution

Since the current must go through both resistors, it will experience the total resistance 28 ohm + 21 ohm = 49 ohm.

• The current must therefore be ( 8 Volts)/( 49 Ohms) = .1632 Amps.
• The voltage change associated with .1632 Amps through 28 ohm is ( .1632 Amps)( 28 ohm) = 4.569 Volts
• Through the 21 ohm resistor the change is ( .1632 Amps)( 21 ohm) = 3.427 Volts.
• Note that these individual voltages add up to the original voltage.

Generalized Solution

When we have two resistors in series, the same current must go through both and their potential differences therefore add (if we go across one, a certain amount of work is done per Coulomb of charge; then when we go across the other an additional amount of work is done per Coulomb; the total is the total voltage across the resistors).  It follows that if the total voltage is V and the resistances are R1 and R2, and the current is I = V / (R1 + R2) and the total resistance must therefore be R1 + R2.

• The voltage drop across each resistor is thus I multiplied by the resistance of that resistor:  V1 = I * R1 and V2 - I * R2.