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### Series and Parallel Circuits

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Date: 10/18/2000 at 19:44:15
From: Craig Kirsch
Subject: Current in parallel and series circuits

I need the formulas for calculating current in a series-parallel
circuit. I know IT, RT, and PT, but I need IR1, IR2, and IR3. I think
I'm missing part of the formula.

```

```
Date: 10/19/2000 at 12:32:27
From: Doctor TWE
Subject: Re: Current in parallel and series circuits

Hi Craig. Thanks for writing to Dr. Math.

In order to solve series or parallel circuits, you have to use both
the series (or parallel) circuit rules and Ohm's laws. For
series-parallel circuits, you have to either (a) break the problem
into series parts and parallel parts, then solve each part using the
appropriate rules, or (b) use simultaneous equations involving
Kirchoff's laws and Ohm's laws.

Ohm's laws relate the four basic electrical characteristics (E, I, R,
and P) to each other. They state:

E = I*R   and   P = I*E

From these two equations, with a little algebra, you can solve for any
of the four characteristics (E, I, R or P) given any two. The
formulas, once solved for the desired characteristic, become:

E = I*R         I = E/R          R = E/I     P = I*E
E = P/I         I = P/E          R = P/I^2   P = I^2*R
E = sqrt(P*R)   I = sqrt(P/R)    R = E^2/P   P = E^2/R

Ohm's laws apply to any individual component, any part of a circuit,
or the circuit totals. Remember to only use the voltage across, the
current through, the resistance of and the power dissipated by the
component(s) in question. (You can't, for example, use IT*R2 to find
E2, you must use I2*R2 - of course, if I2 = IT as it is in series...)

Each series circuit rule relates the individual component values to
the circuit totals for one characteristic. They state:

ET = E1 + E2 + E3 + ...
IT = I1 = I2 = I3 = ...
RT = R1 + R2 + R3 + ...
PT = P1 + P2 + P3 + ...

The ellipsis ('...') means to continue for all components. Remember
that these rules apply to series circuits or series parts ONLY - they
cannot be used on parallel circuits or series-parallel circuits.
Sometimes to remember these rules we say, "In series, current stays
the same but voltage divides." What we mean by 'voltage divides' in
this phrase is that the total voltage is split up (or divided) among
the components.

Similarly, each parallel circuit rule relates the individual component
values to the circuit totals for one characteristic. They state:

ET = E1 = E2 = E3 = ...
IT = I1 + I2 + I3 + ...
1/RT = 1/R1 + 1/R2 + 1/R3 + ...   (called the "reciprocal rule")
PT = P1 + P2 + P3 + ...

The rule for resistance - sometimes called the reciprocal rule - is
tricky. You have to take the reciprocal (use the [1/x] key on your
calculator) of each component's resistance, add them up, then take the
reciprocal of the sum to get the total resistance. In a parallel
circuit, the total resistance should always be LESS than any of the
individual component resistances.

Remember that these rules apply to parallel circuits or parallel parts
ONLY - they cannot be used on series circuits or series-parallel
circuits. Sometimes to remember these rules we say, "In parallel,
voltage stays the same but current divides."

I'll provide two simple examples, one series and one parallel, as a
guide. I like to use a "PIER chart," which is simply a table listing
all the characteristics of the components and the circuit totals. The
chart keeps me organized, lets me know what I have and what I'm
missing, and can even give me clues as to how to find the missing
values. (Whenever I'm working across a row of the PIER chart, I have
to use Ohm's laws; whenever I'm working down a column, I have to use
the series or parallel circuit rules.)

Example I: Series Circuit

B represents a voltage source (battery), and R1, R2 and R3 are
components (resistors).

First, I'll make my PIER chart and fill in my known values:

Comp |   E  |  I |  R  |   P
------+------+----+-----+-------
R1  |      |    | 20@ |
------+------+----+-----+-------
R2  |      |    | 10@ |
------+------+----+-----+-------
R3  |      |    | 30@ |
------+------+----+-----+-------
Tl  | 300V |    |     |

(I'm using @ for "Ohms" because the standard character set doesn't
have the omega symbol.)

Looking at my chart, I see that I have all the resistances except the
total, so I can use my series circuit rule to find RT:

RT = R1 + R2 + R3
= 20@ + 10@ + 30@
= 60@

I can put that in my chart:

Comp |   E  |  I |  R  |   P
------+------+----+-----+-------
R1  |      |    | 20@ |
------+------+----+-----+-------
R2  |      |    | 10@ |
------+------+----+-----+-------
R3  |      |    | 30@ |
------+------+----+-----+-------
Tl  | 300V |    | 60@ |

Now I see that I have the total voltage and resistance. I can use
Ohm's laws to find the total current and total power:

IT = ET / RT        PT = ET^2 / RT
= 300V / 60@        = (300V)^2 / 60@
= 5A                = 90000/60 W
= 1500 W

Note that to avoid having to use squares, I could have computed IT
first, then used the formula PT = IT * ET to find the total power.
Putting these values in my PIER chart:

Comp |   E  |  I |  R  |   P
------+------+----+-----+-------
R1  |      |    | 20@ |
------+------+----+-----+-------
R2  |      |    | 10@ |
------+------+----+-----+-------
R3  |      |    | 30@ |
------+------+----+-----+-------
Tl  | 300V | 5A | 60@ | 1500W

Remembering that in a series circuit the current stays the same, I can
copy the 5A in IT up the entire current column. Mathematically:

IT = I1 = I2 = I3
5A = I1 = I2 = I3

So I1 = 5A, I2 = 5A and I3 = 5A. Putting these in my chart:

Comp |   E  |  I |  R  |   P
------+------+----+-----+-------
R1  |      | 5A | 20@ |
------+------+----+-----+-------
R2  |      | 5A | 10@ |
------+------+----+-----+-------
R3  |      | 5A | 30@ |
------+------+----+-----+-------
Tl  | 300V | 5A | 60@ | 1500W

Finally, I can use Ohm's laws to find E1, P1, E2, P2, E3, and P3.

E1 = I1 * R1      P1 = I1^2 * R1
= 5A * 20@        = 5A^2 * 20@
= 100V            = 25*20 W
= 500W

E2 = I2 * R2      P2 = I2^2 * R2
= 5A * 10@        = 5A^2 * 10@
= 50V             = 25*10 W
= 250W

E3 = I3 * R3      P3 = I3^2 * R3
= 5A * 30@        = 5A^2 * 30@
= 150V            = 25*30 W
= 750W

As with the totals, I could calculate current first, then use P = I*E
for each component if I wanted to avoid having to use squares.
Completing my chart:

Comp |   E  |  I |  R  |   P
------+------+----+-----+-------
R1  | 100V | 5A | 20@ |  500W
------+------+----+-----+-------
R2  |  50V | 5A | 10@ |  250W
------+------+----+-----+-------
R3  | 150V | 5A | 30@ |  750W
------+------+----+-----+-------
Tl  | 300V | 5A | 60@ | 1500W

To double-check, the total voltage times the total current should
always equal the sum of all the component power dissipations. This
works on all circuits: series, parallel, series-parallel, or complex.

ET * IT  =?  P1 + P2 + P3
300V * 5A  =?  500W + 250W + 750W
1500W  =?  1500W

So it checks.

Example II: Parallel Circuit

R1 = 15 Ohms, I2 = 1 A, R3 = 10 Ohms, P3 = 90 W

This circuit is a bit trickier because we're given different
information about different components. Here's where a PIER chart
really is useful.

First, I'll make my PIER chart and fill in my known values:

Comp |  E  |  I |  R  |   P
------+-----+----+-----+------
R1  |     |    | 15@ |
------+-----+----+-----+------
R2  |     | 1A |     |
------+-----+----+-----+------
R3  |     |    | 10@ |  90W
------+-----+----+-----+------
Tl  |     |    |     |

Looking at my chart, I see that I have two of the component values for
R3, so I can use Ohm's laws to find E3 and I3:

E3 = sqrt(P3 * R3)       I3 = sqrt(P3 / R3)
= sqrt(90W * 10@)        = sqrt(90W / 10@)
= sqrt(900) V            = sqrt(9) A
= 30V                    = 3A

In this case, I can't avoid using the square root function altogether,
but once I use it to find the voltage, I could use I3 = E3/R3 to find
the current instead. Putting these values in my chart:

Comp |  E  |  I |  R  |   P
------+-----+----+-----+------
R1  |     |    | 15@ |
------+-----+----+-----+------
R2  |     | 1A |     |
------+-----+----+-----+------
R3  | 30V | 3A | 10@ |  90W
------+-----+----+-----+------
Tl  |     |    |     |

Remembering that in a parallel circuit the voltage stays the same, I
can copy the 30V in E3 up the entire voltage column. Mathematically:

ET = E1 = E2 = E3
ET = E1 = E2 = 30V

So ET = 30V, E1 = 30V and E2 = 30V. Putting these in my chart:

Comp |  E  |  I |  R  |   P
------+-----+----+-----+------
R1  | 30V |    | 15@ |
------+-----+----+-----+------
R2  | 30V | 1A |     |
------+-----+----+-----+------
R3  | 30V | 3A | 10@ |  90W
------+-----+----+-----+------
Tl  | 30V |    |     |

Next, I can use Ohm's laws to find I1, P1, R2 and P2:

I1 = E1 / R1       P1 = E1^2 / R1
= 30V / 15@        = 30V^2 / 15@
= 2A               = 900/15 W
= 60W

R2 = E2 / I2      P2 = I2 * E2
= 30V * 1A        = 1A * 30V
= 30@             = 30W

Again, to avoid having to use squares, I could have computed I1 first,
then used the formula P1 = I1 * E1 to find the power. Putting these
values in my PIER chart:

Comp |  E  |  I |  R  |   P
------+-----+----+-----+------
R1  | 30V | 2A | 15@ |  60W
------+-----+----+-----+------
R2  | 30V | 1A | 30@ |  30W
------+-----+----+-----+------
R3  | 30V | 3A | 10@ |  90W
------+-----+----+-----+------
Tl  | 30V |    |     |

Finally, I can use the parallel circuit rules to find IT, RT, and PT.

IT = I1 + I2 + I3     PT = P1 + P2 + P3
= 2A + 1A + 3A        = 60W + 30W + 90W
= 6A                  = 180W

1/RT = 1/R1 + 1/R2 + 1/R3
= 1/15@ + 1/30@ + 1/10@
= 0.0667 + 0.0333 = 0.1   (or 1/RT = 2/30 + 1/30 + 3/30
= 0.2                              = 6/30
RT = 1/0.2 @                       RT = 30/6 @
= 5@                               = 5@)

In this case, I could avoid having to use the reciprocal rule by using
Ohm's law (RT = ET / IT) once I find the total current - but I'm not
always going to be that lucky. Completing my chart:

Comp |  E  |  I |  R  |   P
------+-----+----+-----+------
R1  | 30V | 2A | 15@ |  60W
------+-----+----+-----+------
R2  | 30V | 1A | 30@ |  30W
------+-----+----+-----+------
R3  | 30V | 3A | 10@ |  90W
------+-----+----+-----+------
Tl  | 30V | 6A |  5@ | 180W

Note that the total resistance (5 Ohms) is less than each of the
individual component resistances. To double-check, the total voltage
times the total current should always equal the sum of all the
component power dissipations:

ET * IT  =?  P1 + P2 + P3
30V * 6A  =?  60W + 30W + 90W
180W  =?  180W

So it checks.

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Physics/Chemistry

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