Today’s agenda:

Resistors in Series and Parallel.You must be able to calculate currents and voltages in circuit components in series and in parallel.

Kirchoff’s Rules.You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit components that are not simply in series or in parallel.

Resistances in Circuits

There are “two” ways to connect circuit elements.

Series:

A B

Put your finger on the wire at A. If you can move along the wires to B without ever having a choice of which wire to follow, the circuit components are connected in series.

Truth in advertising: it is possible to have circuit elements that are connected neither in series nor in parallel. See problem 24.73 in the 12th

editionof our text for an example with capacitors.

Parallel:

A B

Put your finger on the wire at A. If in moving along the wires to B you ever have a choice of which wire to follow, the circuit components are connected in parallel.*

*Truth in advertising: actually, the circuit components are not connected in series, and may

be connected in parallel.

???

Are these resistors in series or parallel?

It matters where you put the source of emf.

+ -

V

parallel

Are these resistors in series or parallel?

It matters where you put the source of emf.

+- V

series

If resistors “see” the same potential difference, they are in parallel. If resistors “see” the same current, they are in series.

It’s difficult to come up with a simple one- or two-sentence rule for series/parallel.

+- V

series

+ -

V

parallel

I

V

Here’s a circuit with three resistors and a battery:

R3R2R1

+ -

VI

Current flows…

…in the steady state, the same current flows through all resistors…

III

…there is a potential difference (voltage drop) across each resistor.

V1 V3V2

Applying conservation of energy allows us to calculate the equivalent resistance of the series resistors.

I am including the derivation in these notes, for the benefit of students who want to look at it.

In lecture, I will skip ahead past the derivation.

An electric charge q is given a potential energy qV by the battery.

R3R2R1

+ -

VI

III

V1 V3V2

As it moves through the circuit, the charge loses potential energy qV1 as it passes through R1, etc.

The charge ends up where it started, so the total energy lost must equal the initial potential energy input:

qV = qV1 + qV2 + qV3 .

V = V1 + V2 + V3

V = IR1 + IR2 + IR3

Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in series.

R3R2R1

+ -

VI

III

V1 V3V2

qV = qV1 + qV2 + qV3

Req

+ -

VI

V

I

As above: V = IReq

From before: V = IR1 + IR2 + IR3

Combining: IReq = IR1 + IR2 + IR3

Req = R1 + R2 + R3

For resistors in series, the total resistance is the sum of the separate resistances.

We can generalize this to any number of resistors:

(resistors in series)a consequence of conservation of energy

R3R2R1

+ -

V

eq ii

R R

Note: for resistors in parallel, Req is always greater than any of the Ri.

V

V

V

R3

R2

R1

+ -

VI

Current flows…

…different currents flows through different resistors…

…but the voltage drop across each resistor is the same.

I3

I1

I2

Here’s another circuit with three resistors and a battery.

Applying conservation of charge allows us to calculate the equivalent resistance of the parallel resistors.

I am including the derivation in these notes, for the benefit of students who want to look at it.

In lecture, I will skip ahead past the derivation.

V

V

V

R3

R2

R1

+ -

VI

I3

I1

I2

A B

In the steady state, the current I “splits” into I1, I2, and I3 at point A.

I

I1, I2, and I3 “recombine” to make a current I at point B.

Therefore, the net current flowing out of A and into B is I = I1 + I2 + I3 .

1 2 31 2 3

V V VI = I = I =

R R R

Because the voltage drop across each resistor is V:

Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in parallel.

V

Req

+ -

VI

I

A B

IFrom above, I = I1 + I2 + I3, and

1 2 31 2 3

V V VI = I = I = .

R R R

So thateq 1 2 3

V V V V = + + .

R R R R

Dividing both sides by V gives

eq 1 2 3

1 1 1 1 = + + .

R R R R

We can generalize this to any number of resistors:

(resistors in parallel)a consequence of conservation of chargeieq i

1 1

R R

Note: for resistors in parallel, Req is always less than any of the Ri.

Summary:

Series A Beq i

i

R Rsame I, V’s add

Parallel A B

same V, I’s add

ieq i

1 1

R R

“just like” capacitors

“just like” capacitors

“just like” capacitors NOT

“just like” capacitors NOT