Series Circuits

AKA Voltage Dividers

Cardinal Rules for Series Circuits

•Potential difference is divided proportionately amongst each of the resistors placed in series with the battery. The sum of the drops in potential at each device in the circuit should add the the emf applied to the circuit. (Voltage is constant)

Energy In = Energy Out E = V1 + V2 + … + Vn

• The total current drawn from the battery is the same through any resistor placed in series with the battery.

IT = I1 = I2 = ….. = In

• The total resistance in a circuit is equal to the sum of all the individual resistances placed in series.

RT = R1 + R2 + … + Rn

V I R

Source Totals 9.00 V

r 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Series Circuits

r R1

R2

R3

B

A

E

V I R

Source Totals 9.00 V 60.95Ω

r 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Series Circuits

r R1

R2

R3

B

A

All resistances are added together to provide the RT

=+

++

E

V I R

Source Totals 9.00 V IT = E/RT

0.148 A60.95Ω

r 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Series Circuits

r R1

R2

R3

B

A

E

The total current drawn from the battery is determined by the total resistance.

V I R

Source Totals 9.00 V 0.148 A 60.95Ω

r 0.148 A 0.20Ω

Resistor 1 0.148 A 15.00Ω

Resistor 2 0.148 A 25.25Ω

Resistor 3 0.148 A 20.50Ω

Series Circuits

r R1

R2

R3

B

A

E

The total current drawn from the battery flows through every part of this circuit.

I is constant

==

==

A = 0.148A

AA = 0.148A

A = 0.148A

A = 0.148A

V I R

Source Totals 9.00 V 0.148 A 60.95Ω

r 0.148 A 0.20Ω

Resistor 1 0.148 A 15.00Ω

Resistor 2 0.148 A 25.25Ω

Resistor 3 0.148 A 20.50Ω

Series Circuits

r R1

R2

R3

B

A

E

V I R

Source Totals 9.00 V 0.148 A 60.95Ω

r 0.030 V 0.148 A 0.20Ω

Resistor 1 2.22 V 0.148 A 15.00Ω

Resistor 2 3.74 V 0.148 A 25.25Ω

Resistor 3 0.148 A 20.50Ω

Series Circuits

rR1

R2

R3

B

A

E

As electric current flows across each resistor it spends energy & those moving charges experience a drop in their potential (voltage drop)

V = 0.030 V

V = 3.74 V V

V = 2.22 V

=

=

=

x

x

x

V I R

Source Totals 9.00 V 0.148 A 60.95Ω

r 0.030 V 0.148 A 0.20Ω

Resistor 1 2.22 V 0.148 A 15.00Ω

Resistor 2 3.74 V 0.148 A 25.25Ω

Resistor 3 3.03 V 0.148 A 20.50Ω

Series Circuits

r R1

R2

R3

B

A

E

V3 could be calculated the same way….. = x

V I R

Source Totals 9.00 V 0.148 A 60.95Ω

r 0.030 V 0.148 A 0.20Ω

Resistor 1 2.22 V 0.148 A 15.00Ω

Resistor 2 3.74 V 0.148 A 25.25Ω

Resistor 3 3.01 V 0.148 A 20.50Ω

Series Circuits

r R1

R2

R3

B

A

E

…. Or V3 could be the remaining difference in potential.

=-

--

V I R

Source Totals 9.00 V 0.148 A 60.95Ω

r 0.030 V 0.148 A 0.20Ω

Resistor 1 2.22 V 0.148 A 15.00Ω

Resistor 2 3.74 V 0.148 A 25.25Ω

Resistor 3 3.01 V 0.148 A 20.50Ω

Series Circuits

r R1

R2

R3

B

A

E

Terminal voltage or VAB is the energy supplied by the battery that is supplied to and used in the circuit.

VAB= E – IT r

VAB= 9.00 V – 0.030 V = 8.97 V

Less = VAB

Parallel Circuits

AKA Current Dividers

Cardinal Rules for Parallel Circuits

• Potential difference is the same across all parallel paths (Voltage is constant)E = V1 = V2 = ….. = Vn

• The total current drawn from the battery is divided proportionately amongst each of the parallel paths

IT = I1 + I2 + ….. + In

• One resistor having the same resistance as all resistance offered by the load on the parallel paths is referred to as the equivalent resistance. A similar idea to total resistance in series circuits. The equivalent resistance is equal to the reciprocal of the sum of the reciprocals of those resistors in parallel.

Req = _______1__________ 1/R1 + 1/R2 + … + 1/Rn

V I R

Source Totals 9.00 V

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

R1

R2

R3E

These are the same 3 resistors just arranged in parallel.

Parallel Circuits

R1

R2

R3E

First, we need to solve for Req

Req = _______1__________ 1/R1 + 1/R2 + 1/R3

= _________1_________

1/15.0Ω + 1/25.25Ω + 1/20.50Ω

= 6.45 Ω

NOTE: The value of Req is always smaller than the smallest resistor in parallel.

V I R

Source Totals 9.00 V IT = E/Req

1.40 A6.45Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

R1

R2

R3E

Using Ohm’s Law still calculates the total current drawn from the battery.

A = IT

V I R

Source Totals 9.00 V 1.40 A 6.45Ω

Resistor 1 9.00 V 15.00Ω

Resistor 2 9.00 V 25.25Ω

Resistor 3 9.00 V 20.50Ω

Parallel Circuits

R1

R2

R3E =

==

The emf applied to the circuit is applied across each parallel branch.

V is constant.

V = E

V I R

Source Totals 9.00 V 1.40 A 6.45Ω

Resistor 1 9.00 V I1 = V1/R1

= 0.600 A 15.00Ω

Resistor 2 9.00 V I2 = V2/R2

= 0.356 A 25.25Ω

Resistor 3 9.00 V 20.50Ω

Parallel Circuits

R1

R2

R3E

Using the voltage drops across each resistor & the resistance, current flowing across each of the resistors can be found.

I is divided amongst the parallel branches.

A = I1

A = I2

A = I3

V I R

Source Totals 9.00 V 1.40 A 6.45Ω

Resistor 1 9.00 V I1 = V1/R1

= 0.600 A 15.00Ω

Resistor 2 9.00 V I2 = V2/R2

= 0.356 A 25.25Ω

Resistor 3 9.00 V I3 = V3/R3

= 0.439 A 20.50Ω

Parallel Circuits

R1

R2

R3E

I3 can be found using the exact same method as the other currents or…..

V I R

Source Totals 9.00 V 1.40 A 6.45Ω

Resistor 1 9.00 V 0.600 A 15.00Ω

Resistor 2 9.00 V 0.356 A 25.25Ω

Resistor 3 9.00 V 0.444 A 20.50Ω

Parallel Circuits

R1

R2

R3E

…or by finding the remaining current.

IT = I1 + I2 + I3

=-

-

V I R

Source Totals 9.00 V

r 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

How would this change if we put the internal resistor back into the circuit?

V I R

Source Totals 9.00 V

RT = Req + r= 6.45Ω+0.20Ω

= 6.65Ω

r 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

The internal resistor is always treated as a resistor in series with the battery and the Req

Req

V I R

Source Totals 9.00 V

IT = E/Req1.35 A 6.65Ω

r 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

The total resistance still determines the current drawn from the battery.

V I R

Source Totals 9.00 V 1.35 A 6.65Ω

r 1.35 A 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

Since the r is considered to be in series with the battery it receives the IT.

=

V I R

Source Totals 9.00 V 1.35 A 6.65Ω

r Vr = IT x r 0.27 V 1.35 A 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

Ohm’s Law calculates the energy spent moving the current out of the battery (volatge drop).

= x

V I R

Source Totals 9.00 V 1.35 A 6.65Ω

r 0.27 V 1.35 A 0.20Ω

Resistor 1 15.00Ω

Resistor 2 25.25Ω

Resistor 3 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

The VAB is the energy actaully supplied to the circuit AND what is applied across each parallel branch.

=

VAB = E – IT r = 9.00 V – 0.27 V = 8.73 V

V I R

Source Totals 9.00 V 1.35 A 6.65Ω

r 0.27 V 1.35 A 0.20Ω

Resistor 1 8.73 V 15.00Ω

Resistor 2 8.73 V 25.25Ω

Resistor 3 8.73 V 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

The VAB is the energy actaully supplied to the circuit AND what is applied across each parallel branch.

V I R

Source Totals 9.00 V 1.35 A 6.65Ω

r 0.27 V 1.35 A 0.20Ω

Resistor 1 8.73 V I1 = V1/R1

= 0.582 A 15.00Ω

Resistor 2 8.73 V

I2 = V2/R2

= 0.346 A 25.25Ω

Resistor 3 8.73 V

I3 = V3/R3

= 0.426 A 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

Ohm’s Law determines the portion of the total current flowing down each parallel branch.

V I R

Source Totals 9.00 V 1.35 A 6.65Ω

r 0.27 V 1.35 A 0.20Ω

Resistor 1 8.73 V 0.582 A 15.00Ω

Resistor 2 8.73 V 0.346 A 25.25Ω

Resistor 3 8.73 V I3 = IT - (I1+ I2)

= 0.422 A 20.50Ω

Parallel Circuits

r

R1

R2

R3

B

A

E

There is an alternate solution for I3 as it is the remaining current.