Series-Parallel Circuits Problem

2011 April 15
by roma

The following is a problem that highlights one aspect of Ohm's Law that I'm still unclear about:


Source: http://www.electronicstheory.com/COURSES/ELECTRONICS/e101-15.htm


Find the voltage drops for R2 and R3.

Given:
R1 = 20 Ohms
R2 = 10 Ohms
R3 = 10 Ohms
E Total = 25 V

Rtotal = 1 / [(1/20) + (1/20)] = 10 Ohms
Itotal = E/Rtotal = 25V/10 Ohms = 2.5A

Now, in order to proceed, we can simplify the circuit diagram:


Source: http://www.electronicstheory.com/COURSES/ELECTRONICS/e101-15.htm


I Total = I1 + I2 + I3

Because R2 and R3 are in series, they have equal currents.
Thus,

I Total = I1 + I2

The resistances for R1 (20 Ohms) and R2+3 (20 Ohms) are the same. That means that the currents are equal.

I1 = Itotal / 2 = 2.5A / 2 = 1.25 A

I do not understand how I would calculate this current if the resistance for R2+3 was not equal to R1.

I1 = 1.25 A
I2 = 1.25 A
I3 = 1.25 A

E1 = I1R1
= (1.25 A) (20 Ohms)
= 25 V

E2 = I2R2
= (1.25 A) (10 Ohms)
= 12.5 V

E3 = I3R3
= (1.25 A) (10 Ohms)
= 12.5 V

This example illustrates how voltages add up for components in series.

One Response leave one →
  1. Tony Abboud permalink
    April 20, 2011

    Hello, I saw your question on open study and decided to check it out.

    Well I see what is puzzling you and you are just forgetting one little thing. That the voltage across resistors in parallel are equal. So in this case V across R1 = 25 volts = Voltage across R2+3

    So, in this case since they are both equal:
    I = V/R
    I1 = 25v / 20 ohm = 1.25 amps
    I2+3 = 25v / 20 ohm = 1.25 amps

    However, lets say R2+3 = 15 ohms (just theoretically speaking since this would change the whole problem)
    but, I2+3 = 25v / 15 ohms = 1.66 amps.

    Hope I helped out. Feel free to email me if you have any other questions.

    -Tony

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