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General Physics (PHY 2140)

Lecture 10Lecture 10 Electrodynamics

Direct current circuits parallel and series connections Kirchhoff’s rules

Chapter 18

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Last lecture:

1.1. Current and resistanceCurrent and resistance

Temperature dependence of resistanceTemperature dependence of resistance

Power in electric circuitsPower in electric circuits

QI

t

1o oR R T T 2

2 VP I V I R

R

Review Problem: Consider a moose standing under the tree during the lightning storm. Is he ever in danger? What could happen if lightning hits the tree under which he is standing?

A branch: A branch is a single electrical element or device (resistor, etc.).

A junction: A junction (or node) is a connection point between two or more branches.

If we start at any point in a circuit (node), proceed through connected electric devices back to the point (node) from which we started, without crossing a node more than one time, we form a closed-path (or loop).

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A circuit with 5 branches.

A circuit with 3 nodes.

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Steady current (constant in magnitude and direction)• requires a complete circuit• path cannot be only resistance

cannot be only potential drops in direction of current flow

Electromotive Force (EMF)• provides increase in potential • converts some external form of energy into electrical energy

Single emf and a single resistor: emf can be thought of as a “charge pump”

I

V = IR

+ - V = IR =

Each real battery has some internal resistance

AB: potential increases by on the source of EMF, then decreases by Ir (because of the internal resistance)

Thus, terminal voltage on the battery V is

Note: is the same as the terminal voltage when the current is zero (open circuit)

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r

R

A

B C

DV Ir E

Now add a load resistance R Since it is connected by a

conducting wire to the battery → terminal voltage is the same as the potential difference across the load resistance

Thus, the current in the circuit is

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r

R

A

B C

D

IR r

E

,V Ir IR or

Ir IR

E

E

Power output:

2 2I I r I R E

Note: we’ll assume r negligible unless otherwise is stated

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Voltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device.

V

Ammeters measure current through a device by being placed in series with the device.

A

Two Basic Principles: Conservation of Charge Conservation of Energy

Resistance Networks

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ab eq

abeq

V IR

VR

I

Req

Ia

b

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1 21 2

eq

eq

V IR

IR IRVR R R

I I

1 2I I I

1. Because of the charge conservation, all charges going through the resistor R2 will also go through resistor R1. Thus, currents in R1 and R2 are the same,

1 2V IR IR

2. Because of the energy conservation, total potential drop (between A and C) equals to the sum of potential drops between A and B and B and C,

By definition,

Thus, Req would be

1 2eqR R R

R 2

R 1

v 2

v 1+ +

+

_

_

_v i1

A B

C

I

Analogous formula is true for any number of resistors,

It follows that the equivalent resistance of a series combination of resistors is greater than any of the individual resistors

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1 2 3 ...eqR R R R (series combination)

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R 2

R 1

v 2

v 1+ +

+

_

_

_v i1

A B

C

I

In the electrical circuit below, find voltage across the resistor RIn the electrical circuit below, find voltage across the resistor R11 in terms of in terms of

the resistances Rthe resistances R11, R, R22 and potential difference between the battery’s and potential difference between the battery’s

terminals V. terminals V.

1 2V V V Energy conservation implies:

with 1 1 2 2andV IR V IR

Then, 1 21 2

, soV

V I R R IR R

Thus,1

11 2

RV V

R R

This circuit is known as voltage divider.

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1 2

1 2 1 2

eq

eq

VI

R

V VV V VI

R R R R R

1 2V V V

1. Since both R1 and R2 are connected to the same battery, potential differences across R1 and R2 are the same,

1 2I I I

2. Because of the charge conservation, current, entering the junction A, must equal the current leaving this junction,

By definition,

Thus, Req would be

1 2

1 1 1

eqR R R

I

I 2 I 1

R 2 R 1V

+

_

I

R e qV

+

_

A

or 1 2

1 2eq

R RR

R R

Analogous formula is true for any number of resistors,

It follows that the equivalent resistance of a parallel combination of resistors is always less than any of the individual resistors

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(parallel combination)

1 2 3

1 1 1 1...

eqR R R R

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In the electrical circuit below, find current through the resistor RIn the electrical circuit below, find current through the resistor R11 in terms of in terms of

the resistances Rthe resistances R11, R, R22 and total current I induced by the battery. and total current I induced by the battery.

1 2I I I Charge conservation implies:

with 1 21 2

, andV V

I IR R

Then, 1 21

1 1 2

, witheqeq

IR R RI R

R R R

Thus, 21

1 2

RI I

R R

This circuit is known as current divider.

I

I 2 I 1

R 2 R 1

+

_

V

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Find the currents IFind the currents I11 and I and I22 and the voltage V and the voltage Vxx in the circuit shown below. in the circuit shown below.

I1I2

4 1 2 2 0 V

7

+_

+

_

V x

IStrategy:

1. Find current I by finding the equivalent resistance of the circuit

2. Use current divider rule to find the currents I1 and I2

3. Knowing I2, find Vx.

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Find the currents IFind the currents I11 and I and I22 and the voltage V and the voltage Vxx in the circuit shown below. in the circuit shown below.

I1I2

4 1 2 2 0 V

7

+_

+

_

V x

IFirst find the equivalent resistance seen by the 20 V source:

4 (12 )7 10

12 4eqR

Then find current I by,20 20

210eq

V VI A

R

We now find I1 and I2 directly from the current division rule:

1 2 1

2 (4 )0.5 , and 1.5

12 4

AI A I I I A

Finally, voltage Vx is 2 4 1.5 4 6xV I A V

• The procedure for analyzing complex circuits is based on the principles of conservation of charge and energy

• They are formulated in terms of two Kirchhoff’s rules:

1. The sum of currents entering any junction must equal the sum of the currents leaving that junction (current or junction rule) .

2. The sum of the potential differences across all the elements around any closed-circuit loop must be zero (voltage or loop rule).

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•The sum of the currents entering a node (junction point) equal to the sum of the currents leaving.

I a

I b

I c

I d

I a , I b , I c , a n d I d c a n e a c h b e e ith e r a p o s itiv eo r n e g a tiv e n u m b e r.

Ia + Ib = I c + Id

11

As a consequence of the Law of the conservation of charge, we have: As a consequence of the Law of the conservation of charge, we have:

Similar to the water flow in a pipe.

1. Assign symbols and directions of currents in the loop– If the direction is chosen wrong, the current will come out with a

right magnitude, but a negative sign (it’s ok).2. Choose a direction (cw or ccw) for going around the loop.

Record drops and rises of voltage according to this:– If a resistor is traversed in the direction of the current: +V = +IR– If a resistor is traversed in the direction opposite to the current: -

V=-IR– If EMF is traversed “from – to + ”: +– If EMF is traversed “from + to – ”: -

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•The sum of the potential differences across all the elements around any closed loop must be zero.

11

As a consequence of the Law of the conservation of energy, we have: As a consequence of the Law of the conservation of energy, we have:

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Loops can be chosen arbitrarily. For example, the circuit below contains a number of closed paths. Three have been selected for discussion.

+

+

+

+ +

+

+

+

+

+

+

-- -

-

-

-

--

-

-

-v1

v2

v4

v3

v12

v11 v9

v8

v6

v5

v7

v10

+

-

Path 1

Path 2

Path 3

Suppose that for each element, respective current flows from + to - signs.

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+

+

+

+ +

+

+

+

+

+

+

-- -

-

-

-

--

-

-

-v1

v2

v4

v3

v12

v11 v9

v8

v6

v5

v7

v10

+

-

“a”•

Blue path, starting at “a”

- v7 + v10 – v9 + v8 = 0

•“b”

Red path, starting at “b”

+v2 – v5 – v6 – v8 + v9 – v11

– v12 + v1 = 0

Yellow path, starting at “b”

+ v2 – v5 – v6 – v7 + v10 – v11

- v12 + v1 = 0

Using sum of the drops = 0

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Example: For the circuit below find I, V1, V2, V3, V4 and the power supplied by the 10 volt source.

1. For convenience, we start at point “a” and sum voltage drops =0 in the direction of the current I.

+10 – V1 – 30 – V3 + V4 – 20 + V2 = 0 (1)

+ +

+

+

+

+

+_ _

_

_

_

_

_V 1

V 4

V 3 V 2

3 0 V 1 0 V

1 5 4 0

5

2 0

2 0 V

I

" a"

2. We note that: V1 = - 20I, V2 = 40I, V3 = - 15I, V4 = 5I (2)

3. We substitute the above into Eq. 1 to obtain Eq. 3 below.

10 + 20I – 30 + 15I + 5I – 20 + 40I = 0 (3)

Solving this equation gives, I = 0.5 A.

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Using this value of I in Eq. 2 gives:

V1 = - 10 V

V2 = 20 V

V3 = - 7.5 V

V4 = 2.5 V

P10(supplied) = -10I = - 5 W

(We use the minus sign in –10I because the current is entering the + terminal)In this case, power is being absorbed by the 10 volt supply.

+ +

+

+

+

+

+_ _

_

_

_

_

_V 1

V 4

V 3 V 2

3 0 V 1 0 V

1 5 4 0

5

2 0

2 0 V

I

" a"

Consider the circuit

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i

RC

switch

+q -q

vc = q/C vR = iR

t q i0 0 1

0.02 0.064493015 0.9355070.04 0.124826681 0.87517330.06 0.181269247 0.81873080.08 0.234071662 0.7659283

0.1 0.283468689 0.71653130.12 0.329679954 0.670320.14 0.372910915 0.62708910.16 0.41335378 0.58664620.18 0.451188364 0.5488116

0.2 0.486582881 0.51341710.22 0.519694699 0.48030530.24 0.550671036 0.4493290.26 0.579649615 0.42035040.28 0.606759279 0.3932407

0.3 0.632120559 0.36787940.32 0.655846213 0.34415380.34 0.678041728 0.32195830.36 0.698805788 0.30119420.38 0.718230711 0.2817693

0.4 0.736402862 0.2635971

0 0.2 0.4 0.6 0.8 1 1.2

q

t

i

1 t RCq Q e RC is called the time constant

Discharge the capacitor

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i

RC

switch

+q -q

vc = q/C vR = iR

t RCq Qe

t q i0 1 1

0.02 0.935506985 0.9355070.04 0.875173319 0.87517330.06 0.818730753 0.81873080.08 0.765928338 0.7659283

0.1 0.716531311 0.71653130.12 0.670320046 0.670320.14 0.627089085 0.62708910.16 0.58664622 0.58664620.18 0.548811636 0.5488116

0.2 0.513417119 0.51341710.22 0.480305301 0.48030530.24 0.449328964 0.4493290.26 0.420350385 0.42035040.28 0.393240721 0.3932407

0.3 0.367879441 0.36787940.32 0.344153787 0.34415380.34 0.321958272 0.32195830.36 0.301194212 0.30119420.38 0.281769289 0.2817693

0.4 0.263597138 0.2635971

0 0.2 0.4 0.6 0.8 1 1.2

q

t

i