# Circuits Wired Partially in Series and Partially in Parallel

#### runfast220

Determine the power dissipated in the 5.0 ohms resistor in the circuit shown in the drawing.

see the attachment for the diagram.

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#### werehk

PHF Hall of Fame
For resistors in series R1 and R2, the equivalent resistance(R) ,
$$\displaystyle R= R1 + R2$$

For resistors in parallel R1 and R2, the equivalent resistance(R),
$$\displaystyle \frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}$$

First, group the resistors 2ohms and 1 ohms, 5ohms and 1ohms which are in series into single resistor first$$\displaystyle R_X$$ $$\displaystyle R_Y$$. Then Group the two requvalent resistors $$\displaystyle R_X$$ and $$\displaystyle R_Y$$which are in parallel into single resistor. Two resistors remain in series. Use the formula again.

$$\displaystyle R_X=5+1$$
$$\displaystyle R_Y=2+1$$
Their equivalent resistance $$\displaystyle R_Z$$
$$\displaystyle \frac{1}{R_Z} = \frac{1}{R_X}+\frac{1}{R_Y}$$

R_Z + 4 would be the final equivalent resistance

Use V= IR to find the total current, then find the current through the 5ohms resistor. Power dissipated= (current)^2 X Resistance of resistor

#### sa-ri-ga-ma

Determine the power dissipated in the 5.0 ohms resistor in the circuit shown in the drawing.

see the attachment for the diagram.
In the problem neither the voltage of the source nor the current in any branch is given.
How to determine the power dissipation?

#### runfast220

In the problem neither the voltage of the source nor the current in any branch is given.
How to determine the power dissipation?[/quote

Sorry I forgot to put that in the diagram, the voltage is 12V.

#### physicsquest

PHF Helper
The required power dissipation is = (5 x V^2) / 18 x 18where V is the voltage of the source.

#### runfast220

$$\displaystyle R_X=5+1$$
$$\displaystyle R_Y=2+1$$
Their equivalent resistance $$\displaystyle R_Z$$
$$\displaystyle \frac{1}{R_Z} = \frac{1}{R_X}+\frac{1}{R_Y}$$

R_Z + 4 would be the final equivalent resistance

Use V= IR to find the total current, then find the current through the 5ohms resistor. Power dissipated= (current)^2 X Resistance of resistor[/quote]

so...

1/Rz = (1/6)+(1/3)
Rz=2
2+4=6 final equivalent resistance(why do we add 4?)

I=V/R=12/6 =2

Then I'm lost after that.
How do I find the current through the resistor? is it just: I=V/R=12/5
and then what current and resistance do I use in the final equation?

The correct answer is supposed to be 2.2 W.

#### werehk

PHF Hall of Fame
You add 4ohms because the total current(I) is to be found by grouping all resistors into 1 resistor.

Then $$\displaystyle 4I$$ would be the voltage of the 4ohms resistor.
The voltage across $$\displaystyle R_X$$ and $$\displaystyle R_Y$$would be $$\displaystyle 12-4I$$ as they are in parallel, so voltages are the same

So current$$\displaystyle I_1$$across $$\displaystyle R_Y$$ is $$\displaystyle (12-4I)/(R_Y)$$
Then the power dissipitated would be$$\displaystyle I_1^2$$X$$\displaystyle 5$$

runfast220
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