DC Circuits

Resistors in Series and Parallel

resistor:

resistors connected in series:

battery:+ -

V

R

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.

A resistor is any circuit element that has electrical resistance (heater, light bulb, etc.). Usually we assume wires have no resistance.

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

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 .

R3R2R1

+ -

VI

III

V1 V3V2

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.

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 make an OSE:

OSE: Req = Ri (resistors in series)

resistors connected in parallel:

A B

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

a consequence of conservation of energy

V

V

V

R3

R2

R1

+ -

VI

current flows

different currents flows through different resistors

the voltage drop across each resistor is the same

I3

I1

I2

Caution: circuits which are drawn to appear very different may be electrically equivalent.

V

V

V

R3

R2

R1

+ -

VI

I3

I1

I2A 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 make an OSE:

OSE: (resistors in parallel)ieq i

1 1 =

R Ra consequence of conservation of charge

Examples

How much current flows from the battery in the circuit shown? What is the current through the 500 resistor?

+ -

12 V

400

500

700

I I

I1

I2

ab

c

What is your stragegy? Step 1—replace the 500 and 700 parallel combination by a single equivalent resistor.

+ -

12 V

400

500

700

I I

I1

I2

ab

c

Woe is me, what to do?

I = ?

I1 = ?

Woe is me, what to do? Always think: bite-sized chunks!

Step 2—replace the 400 and Req1 series combination by a single equivalent resistor Req, net.

+ -

12 V

400 Req1

I I

ab

c

Woe is me, what to do?

I = ?

I1 = ?

Woe is me, what to do? Find another bite-sized chunk!

+ -

12 V

Req1, net

I I

a c

Step 3—Solve for the current I.

This isn’t so complicated!

Step 4—To get I1, Calculate Vbc.

+ -

12 V

400

500

700

I I

I1

I2

ab

c

Use Vtotal = Vab + Vbc.

You know Vtotal= V and I so you can get Vab and then Vbc.

Knowing I, Calculate I1. Woe is me! Stuck again!

Vbc

Vab

+ -

12 V

400

500

700

I I

I1

I2

ab

c

The voltage drop across both the 500 and 700 resistors is the same, and equal to Vbc. Use V = IR to get I1 across the 500 resistor.

19.2 EMF and Terminal Voltage

We have been making calculations with voltages from batteries without asking detailed questions about the batteries. Now it’s time to ask those questions.

We introduce a new term – emf – in this section.

Any device which transforms a form of energy into electric energy is called a “source of emf.”

“emf” is an abbreviation for “electromotive force,” but emf does not really refer to force!

The emf of a source is the voltage it produces when no current is flowing.

The voltage you measure across the terminals of a battery (or any source of emf) is less than the emf because of internal resistance.

Here’s a battery with an emf. All batteries have an “internal resistance:”

+ -a b The “battery” is everything

inside the green box.

Hook up a voltmeter to measure the emf:

+ -a b The “battery” is everything

inside the green box.

Getting ready to connect the voltmeter.

You can’t measure voltage without some (however small) current flowing, so you can’t measure emf directly.

Measuring the emf???

+ -a b

The “battery” is everything inside the green box.

As soon as you connect the voltmeter, current flows.

You can only measure Vab.

I

We model a battery as producing an emf, , and having an internal resistance r:

+ -a b The “battery” is everything

inside the green box.

The terminal voltage, Vab, is the voltage you measure with current flowing. When a current I flows through the battery, Vab is related to the emf, , by

abV = ε ± I r .

r

Why the sign? If the battery is delivering current, the V it delivers is less than the emf, so the – sign is necessary.

If the battery is being charged, you have to “force” the current through the battery, and the V to “force” the current through is greater than the emf, so the + sign is necessary.

This will become clear as you work (and understand) problems.

Operationally, you simply include an extra resistor to represent the battery resistance, and label the battery voltage as an emf instead of V (units are still volts).

Example

For the circuit below, calculate the current drawn from the battery, the terminal voltage of the battery, and the current in the 6 resistor.

5 4

8

10

6

0.5 = 9 V

The following is a “conceptual” solution. Please go back and put in the numbers for yourself.

In the next section, we will learn a general technique for solving circuit problems. For now, we break the circuit into manageable bits. “Bite-sized chunks.”

Replace the parallel combination by its equivalent.

5 4

8

10

6

0.5 = 9 V

Do you see any bite-sized chunks that are simple series or parallel?

5 4

8

10

6

0.5 = 9 V

Replace the series combination by its equivalent.

Any more “bite-sized chunks?” Pretend that everything inside the green box is a single resistor.

5 4

8

10

6

0.5 = 9 V

You are left with an equivalent circuit of 3 resistors in series, which you can handle.

Next bite-sized chunk. Inside the blue box is “a” resistor. Replace the parallel combination by its equivalent.