S2E4: Series and Parallel

In the network shown you are given that V and that the resistors have the resistances , , , and .

What is the equivalent resistance (in Ohms) of the series combination of and ? 

Ra = R3 + R4

Ra = 2.0 + 2.0 = 4.0 Ohms

What is the equivalent resistance (in Ohms) of the parallel combination of and the series combination of and ?

Rb = (R2 * Ra) / (R2 + Ra)

Rb = (4.0 * 4.0) / (4.0 + 4.0) = 16.0 / 8.0 = 2.0 Ohms

  

What is the equivalent resistance (in Ohms) of the series combination of and the parallel combination of and the series combination of and ? 

Rc = R1 + Rb

Rc = 4.0 + 2.0 = 6.0 Ohms

What is the current (in Amperes) ? 

V = I*R

I = V / R

i1 = 2.0 / 6.0 = 1 / 3 = 0.33 Amps

What is the voltage (in Volts) across the resistor ? 

So create the KVL equations for the loops:

-V + i1*R1 + i2*R2 = 0

-i2*R2 + i3*R3 + i4*R4 = 0

-V + i1*R1 + i3*R3 + i4*R4 = 0

And create the KCL equation for the nodes:

-i1 + i2  + i3 = 0

-i2 + i1- i3 = 0

And we solve in terms of i2:

-V + i1*R1 + i2*R2 = 0

i1*R1 + i2*R2 = V 

i2*R2 = V - (i1*R1)

i2 = (V - (i1*R1)) / R2

i2 = (2.0 - (0.33 * 4.0)) / 4.0 = (2.0 - (1.32)) / 4.0 = 0.68 / 4.0 = 0.17 Amps

And now we use Ohm's Law:

V=IR

v2 = i2 * R2

v2 = 0.17 * 4.0 = 0.68 Volts

What is the voltage (in Volts) across the resistor ? 

We've got some KVL equations that involve i4, but we also need i3.

Fortunately, we've got some KCL equations that give us i3 in terms of i1 and i2.

-i1 + i2 + i3 = 0

i2 + i3 = i1

i3 = i1 - i2

i3 = 0.33 - 0.17 = 0.16

So now we can solve for i4 in the KVL equation:

-i2*R2 + i3*R3 + i4*R4 = 0

i3*R3 + i4*R4 = i2*R2

i4*R4 = i2*R2 - i3*R3

i4 = (i2*R2 - i3*R3) / R4

i4 = ((0.17 * 4.0)  - (0.16 * 2.0)) / 2.0

i4 = ((0.68)  - (0.32)) / 2.0 = 0.36 / 2.0 = 0.18 Amps

And now we can apply Ohm's Law to figure out the voltage:

V = IR

v4 = 0.18 * 2.0

v4 = 0.36 Volts

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This answer was rejected by the courseware, so I went back and looked at the picture again.

The voltage across R2 is the same at the voltage across the combination of R3 and R4, since they are in parallel to one another.

So, figuring out the voltage drop across R4 is then the ratio of R3 to R4. Since R3 = R4, they must split the voltage drop equally, so we take the voltage v2 (0.68) and divide it in half: (0.34)

And 0.34 is the correct answer.

This is an important circuit structure called a ladder. Many filters are built with ladders of capacitors and inductors. We will see this kind of circuit later.