19.1

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The Formulas:


R = \frac{V}{I}
R=Resistance (ohms)
V=Voltage (Volts)
I=Current (Amps)

P =VI = I^2R = \frac{V^2}{R}
R=Resistance (ohms)
V=Voltage (Volts)
I=Current (Amps)
P=Power (Watts)

Series Circuit


19.1.1.GIF

Begin with finding the current:


I=V/R
I=13/(2.4+1.3+6.1+3.4)
I=.9848
store the current, .9848A, in A on your calculator

To find the current running through the ammeters notice that the current can only flow one way through the battery.
Therefore, the current through both ammeters is the same as the current flowing through the entire battery.
A1=.9848A or about .985A
A2=.9848A or about .985A

To find V1:


V=(I)(R)
V1=?
I=A=.9848A
R=2.4ohms

V1=(A)(2.4)
V1=2.36V

To find V2:


V=(I)(R)
V2=?
I=A=.9848A
R=6.1+3.4=9.5ohms

V2=(A)(9.5)
V2=9.356V

To find V3:


V=(I)(R)
V3=?
I=A=.9848A
R=0 ohms (there is not a resistor in this chunk of the battery, therefore the resistance is 0 ohms)

V3=(A)(0)
V3=0V

To find the greatest power dissipated by a resistor:


P=(I^2)(R)
P=?
I=A=.9848A
R= 6.1 ohms (on a series circuit to find the greatest power you use the circuit with the greatest resistance and to find the least power you use the circuit with the smallest resistance)

P=(A^2)(6.1)
P=5.9165 or about 5.92W




Parallel Circuit


19.1.2.GIF
First off, you need to find the currents flowing through each piece of the parallel circuit. The different sections of the circuit are defined with the different colors of current flowing through each resistor; blue, green, and red.

Let's start with the blue current (I1)


I=V/R
I1=?
V=45V
R=2.1 ohms

I1=45/2.1
I1=21.428A (store I1 in alpha A)

Next is the green current (I2)


I=V/R
I2=?
V=45V
R=5.6 ohms

I2=45/5.6
I2=8.0357A (store I2 in alpha B)

Last is the red current I3


I=V/R
I3=?
V=45V
R=4.1 ohms

I3=45/4.1
I3=10.975A (store I3 in alpha C)


The next step is to determine what currents flow through each meter. This is really easy with the lines of color because if the line passes through the meter you add those currents together and you will have the total current through that meter.

Let's start with Meter A1:


The blue, green, and red currents flow through A1, therefore add all of the currents you just found (I1, I2, and I3) together.

A1=A+B+C (keep in mind that you stored the currents in these letters, so use the letters to keep the number accurate, plus it's easier!)
A1=40.439 or about 40.4A

Next is meter A2:


The green and red currents flow through A2. Therefore add the currents you just found for I2 (green) and I3 (red).

A2=B+C
A2=19.011 or about 19.0A

Last is meter A3:


If you look at A3 there is only the green (I3) current flowing through the meter. Therefore, the current for I3 is your answer.
A3=C
A3=10.975A or about 11.0A

Least Power Dissipated by a Resistor


P=(V^2)/R
P=?
V=45V
R=5.6 ohms (for parallel circuits, to find the least power you use the greatest resistor, to find the greatest power you use the smallest resistor)
P=(45^2)/5.6
P=361.6 or about 362W