# Series-Parallel Circuits Problem

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 R_{2} and R_{3}.

Given:

R_{1} = 20 Ohms

R_{2} = 10 Ohms

R_{3} = 10 Ohms

E Total = 25 V

R_{total} = 1 / [(1/20) + (1/20)] = 10 Ohms

I_{total} = E/R_{total} = 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 = I_{1} + I_{2} + I_{3}

Because R_{2} and R_{3} are in series, they have equal currents.

Thus,

I Total = I_{1} + I_{2}

The resistances for R_{1} (20 Ohms) and R_{2+3} (20 Ohms) are the same. That means that the currents are equal.

I_{1} = I_{total} / 2 = 2.5A / 2 = 1.25 A

I do not understand how I would calculate this current if the resistance for R_{2+3} was not equal to R_{1}.

I_{1} = 1.25 A

I_{2} = 1.25 A

I_{3} = 1.25 A

E_{1} = I_{1}R_{1}

= (1.25 A) (20 Ohms)

= 25 V

E_{2} = I_{2}R_{2}

= (1.25 A) (10 Ohms)

= 12.5 V

E_{3} = I_{3}R_{3}

= (1.25 A) (10 Ohms)

= 12.5 V

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

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