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PHYS202 – SPRING 2009Lecture notes – Electric Circuits

Batteries

• A battery is a device that provides a potential difference to two terminals.

• Different metals in an electrolyte will create a potential difference, and maintain the difference so long as the load on the battery is not to large.

• Chemical potential energy is converted to electric potential energy.

• The battery is said to provide Electromotive Force (E)

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Ohm’s Law

• Let us consider a circuit.

• Connections are shown as solid lines.

• Batteries are shown as two lines perpendicular to connections with one shorter than the other.

• The conventional current flows from the positive to negative terminals.

VI ‐ current

R ‐ resistance

Positive

Negative

Ohm’s Law

• Normally current is not labeled.

• The current flow is proportional to the voltage:

• Resistance is the constant of proportionality: Ohm’s Law

•• An element within an electric circuit with resistance is called a resistor.

• Ohm’s Law is not a fundamental rule of nature.

• It isn’t true in every case, everywhere – but it is a very useful rule in practical situations.

VI ‐ current

R ‐ resistance

V I∝

V IR=

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Example20.4 – Suppose that the resistance between the walls of a biological cell is 5.0 x 109 Ω.

(a) What is the current when the potential difference between the walls is 75 mV?

Use Ohm’s law:

(b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.50 s?

119

0.075V 1.5 10 A5.0 10

VIR

V IR

I −

⇒ =

= = ×× Ω

=

( )( )117

19

1Na Na1

1.5 10 A 0.50sNa 4.7 10 Na

1.602 10 C

QI Q I tt

I tN Q Ne e

N ions

++

−+ +

−

Δ= ⇒ Δ = ΔΔ

Δ= Δ × ⇒ =

×= = ×

×

Resistance

• Resistance is given in terms of an objects tendency to resist the flow of electrons ρ.

• Resistance increases with the length of the material and decreases with its cross sectional area.

• All materials have some resistivity, ρ, to differing degrees.

RALρ

=

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Resistance

• As expected, insulators have very high resistivity and conductors very low resistivity.

• Semi‐conductors fall in between and their resistivity depends heavily on the amount of impurities.

Resistance and Temperature

• As temperature changes, so does resistivity.

• In small temperature ranges, this is given by

• Where the ρ0 is the resistivity at temperature T0, and α is the temperature coefficient of resistivity.

( )0 01 T Tρ ρ α⎡ ⎤= + −⎣ ⎦

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Example

• Problem 20.14 – A copper wire has a resistance of 38.0 Ω at 25 ˚C and 43.7 Ω at 55 ˚C. What is the temperature coefficient of resistivity?

• Solving for α

( )–10

0

43.7–1 –138.0 0.0050 C

– 55 C – 25 C

RRT T

α

ΩΩ= = = °

° °

( )0 01 T Tρ ρ α⎡ ⎤= + −⎣ ⎦

Electric Power

The amount of power used in a portion of a circuit, is the product of the voltage drop and the current in that section of the circuit.

Using Ohm’s Law: ,

Energy qV qP P Vt t t

P IV

Δ Δ Δ= =

Δ Δ⇒ =

Δ=

V IR=

22

( ) VP I IR P VR

VP I R PR

⎛ ⎞= = ⎜ ⎟⎝ ⎠

= =

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Example

20.26 – A piece of Nichrome wire has a radius of 6.5 x 10‐4 m. It is used in a laboratory to make a heater that uses 4.00 x 102 W of power when connected to a voltage source of 120 V. Ignoring the effect of temperature on resistance, estimate the necessary length of wire.Looking up the resistance of NiChrome wire, we find ρ = 100 x 10‐8 Ω∙m. We need to find the resistance of a wire with a particular radius:

We need the resistance needed to dissipate 400W by a voltage of 120V:

Thus, the length of the wire must be

22 VP

VPR

R⇒ ==

2

L LRA rρ ρ

π= =

( ) ( )( )( )

2 2 2

22

8

2

4120 V 6.5 10 m48m

100 10 m 400 W

V V rLr

LP P

L

ρπ

πρ

π −

−

= ⇒ =

×= =

× Ω⋅

Example

20.27 – Tungsten has a temperature coefficient of resistivity of 0.0045 (C°)−1. A tungsten wire is connected to a source of constant voltage via a switch. At the instant the switch is closed, the temperature of the wire is 28°C, and the initial power delivered to the wire is P0. At what wire temperature will the power that is delivered to the wire be decreased to ½ of P0?If the source of the voltage is constant and temperature affects resistance, can

compe the power we find , thus what we are looking for is

the point where the resistance is twice the value at 28˚C. Since the diameter and length of the wire are unchanged, we are simply looking for where the resistivity is doubled:

2 2

0 01and2 2

V VP PR R

= =

( ) ( )

( )

0 0 0 0

0 1

2 1 1

1 1 28 C 250 C0.0045 C

T T T T

T T

ρ ρ α α

α −

⎡ ⎤= + − − =⎣ ⎦

= + = =

⇒

+

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Alternating Current

• Let us treat the voltage source as oscillatingas a sine wave (see image to right)

• Normal AC voltage in the US has afrequency of 60 Hz (e.g. most of Europe is50 Hz).

• So what is the current? ‐ Ohm’s Law gives

• The current oscillates in the same way as the voltage.

( )0 sin 2V V ftπ=

60 Hzf =

( )

( )

0

0

sin 2

sin 2

VVI ftR R

I I ft

π

π

= =

=

AC Power

• Power is

• Average power is half of thepeak power.

• Average Power is

( )20 0 sin 2P VI V I ftπ= =

Average Power

0 0 0 0

2 2 2V I V IP ⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

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AC Power

• We define this level of voltage and current as the “root mean square” (rms) voltage and current.

• Since this is the average power delivered over one period, it equates to the power delivered in an AC circuit.

• We apply Ohm’s law V = IR

• In the US, we have Vrms = 120V (240V for heavy duty), which equates to a maximum voltage of 170V (340V)

22

rms rms

rmsrms

P V I

VP P I RR

=

= =

0 0

2 2rms rmsV IV I= =

Example

20.36 – A light bulb is connected to a 120.0V wall socket. The current in the bulb depends on the time t according to the relation I = (0.707 A) sin [(314 Hz)t].(a) What is the frequency of the alternating current?

We know that I = I0sin(2πft), so,

(b) Determine the resistance of the bulb’s filament.

Using Ohm’s law,

(c) What is the average power delivered to the light bulb?

314 Hz 50 Hz2

fπ

= =

( )22 120V60 W

240r

rms rmm

ssVR

P V I= = = =Ω

( )0 / 2

2 120 V240

0.707 A

rmrms rms

s

rms

V VRI I

V I R

R

⇒ = =

= =

=

Ω

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AC Power Transmission

• Note that power transmitted through a power line is

• Let us say that 10 km of wire has a resistance of 1.7 Ω.If a power line is to transmit 10kW, is it better to have high voltage or high current?

• When dealing withpower loss through a wire, use P = I2R.

22

rms rms

rmsrms

P V I

VP P I RR

=

= =

Voltage Current Power loss

1 kV 10A 170W (1.7%)

10kV 1A 1.7W (0.017%)

100V 100A 1.7kW (17%)

Power loss in a wire

• Current must stay constant through a wire, or charge would start to build up.

• The Voltage drop is continuous, so we would have to calculate how much power is lost at each point in the wire due to the constant drop in voltage.

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Wiring Resistors in Series

• Take two resistors (R1 and R2) and place them in series.

• Since charge does not build up, the current through both must be the same, so each must experience a voltage drop:

• Resistors in Series add their resistances together to get the equivalent resistance.

VR1 R2

( )1 2 1 2 1 2V V V IR IR I R R= + = + = +

V1 V2

( ) ( )1 2 1 2Series SeriesIR I R R R R R= + ⇒ = +

1 2SeriesR R R= + +

Wiring Resistors in Series

• Power for resistors in series

• Find the current in the circuit:

• The corresponding voltage dropsacross each resistor are:

• The power dissipated is

• The total power dissipated is as expected

V

R1 R21 2

VIR R

=+

V1 V2

1 1 2 2andV IR V IR= =

( ) ( )

2 21 1 2 22 2

1 1 2 22 21 2 1 2

and

and

P I R P I R

V VP R P RR R R R

= =

= =+ +

( ) ( )

2 2 2

1 2 1 22 21 21 2 1 2

V V VP P P R R P VIR RR R R R

= + = + ⇒ = =++ +

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Examples

20.42 – The current in a 47Ω resistor is 0.12 A. This resistor is in series with a 28Ωresistor, and the series combination is connected across a battery. What is the battery voltage? For resistors in series, the current must be the same, so,

( ) ( )( )1 2 0.12 A 47 289 V

V I R RV

= + = Ω+ Ω

=

Examples

20.47 – Two resistances, R1 and R2, are connected in series across a 12‐V battery. The current increases by 0.20 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.10 A when R1 is removed, leaving R2 connected across the battery. Find (a) R1 and (b) R2.Examining the series situation, we find that I1 = Is+0.20A and I2 = Is+0.10A:

Thus,

( )( ) ( )

1 2

1 1 1 2 2 20.20A 0.10 As

s s

V I R R

V I R I R V I R I R

= +

= = + = = +

( ) ( )

( ) ( ) ( )( ) ( )

( ) ( )

2

2 2

2

1 20.20A 0.10A

0.10A 2 0.30A

0.30A 0.020A 2

0.20A0.

0.30A

0.020

20A 0.10A

A

s s

s s s s

s

s s

s s ss s

I

V VR RI I

V VV I I II

I I

II

I I

I

⇒ = +

+ + =

= =+ +

⎡ ⎤= + + +⎢ ⎥

+ +⎢ ⎥

=

⎣ ⎦+

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Examples

20.47 – (cont.)Thus, , so substituting back,

Note that this is not the only method to get to the solutions.

0.1414 AsI =

( ) ( )1 2

1 2

1 2

0.20A 0.10 A12V 12V

0.20A 0.1414A 0.20 A 0.10A 0.1414 A 0.20 A

35 50

s s

s s

V I R V I RV VR R

I I

R R

= + = +

= = = =+ + + +

= Ω = Ω

Wiring Resistors in Parallel

• Take two resistors (R1 and R2) and place them in parallel.

• The battery provides the same voltage to each resistor:

V R1 R2

R2V R1

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Wiring Resistors in Parallel

• Take two resistors (R1 and R2) and placethem in parallel.

• The battery provides the same voltage toeach resistor.

• The currents must add up to the total current through the battery,

• For resistors in parallel, the net resistance is the inverse of the sum of the inverse of the resistances.

V R1 R2

1 21 2 1 2

1 2 1 2

1 1

1 1 1 1 1

parallel parallel

V VI I I VR R R R

V VR R R R R R

⎛ ⎞= + = + = +⎜ ⎟

⎝ ⎠⎛ ⎞

= + ⇒ = +⎜ ⎟⎝ ⎠

I1 I2

1 2

1 1 1

parallelR R R= + +

Wiring Resistors in Parallel

• Power for resistors in parallel:

• Find the current through theresistors:

• Power is

• The total power given from the battery is

V R1 R2

2 2 22

1 21 2 1 2

1 1

parallel

V V VP P P VR R R R R

⎛ ⎞= + = + = + =⎜ ⎟

⎝ ⎠

I1 I2

1 1 2 2P IV P I V P I V= ⇒ = =

1 21 2

V VI IR R

= =

2 2

1 21 2

V VP PR R

= =

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Examples20.52 – The drawing shows a circuit that contains

a battery, two resistors, and a switch. What isthe equivalent resistance of the circuit whenthe switch is (a) open and (b) closed?When the switch is open, no current flows through R2, thus the equivalent resistance is 65.0Ω.

When the switch is closed the resistors are in parallel, so the equivalent resistance is,

What is the total power delivered to the resistors when the switch is (c) open and (d) closed?

( )( )

1 2

1 21 2

1 1 1

65.0 96.038.8

65.0 96.0

parallel

paralle

parallel

l

R RRRR R

R

R R= +

Ω Ω= =

+ Ω

⇒ =+

ΩΩ

( )

( ) ( )

22

12 22 2

1 2

9.00V1.25W

65.0

9.00V 9.00V2.09 W

65.0 96.0

op

cl

VPR

V VPR R

= = =Ω

= + = + =Ω Ω

Examples

20.56 – A cylindrical aluminum pipe of length 1.50 m has an inner radius of 2.00 x 10‐3

m and an outer radius of 3.00 x 10‐3 m. The interior of the pipe is completely filled with copper. What is the resistance of this unit? As the hint in the book suggests, we can treat this as two parallel resistors. We need to find the resistance of each.The cross sectional areas are

Thus the resistances,

Placing the resistances in parallel,

( )2 2 2Cu in Al out inA R A R Rπ π= = −

( )2 2 2Cu Cu Al Al

Cu AlCu in Al out in

L L L LR RA R A R Rρ ρ ρ ρ

π π= = = =

−

( )

( )

( )( )

2 22

2 2 2

32 2 2

1 1 1

1 1 1

1.16 10

out inin

net Cu Al Cu Al

Al in Cu out in

net Cu Al Cu Al

Cu Alnet

Al in Cu out in

R RRR R R L L

R R R

R R R LLR

R R R

ππρ ρ

πρ πρ

ρ ρρ ρ

π ρ ρ−

−= + = +

+ −= + =

= = × Ω+ −

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Resistors in Series and Parallel

• Circuits can be wired with resistors in series and parallel.

• To find the current and power delivered by a battery, create equivalent circuits by combining resistors:

• Each circuit is equivalent from the point of view of the battery.

• Deal with each part of the circuit where you are combining two resistors, then move to the next.

V Rp

1

1 1 1

p SR R R= +

V R1 R3

R2

V R1 R3

2 3SR R R= +

Examples

20.58 – Find the equivalent resistance between pointsA and B in the drawing. We must combine the resistors until only one remains.Starting with the three resistors in series on the right,we find three resistors in series that can be combined:

Now, there are two resistors in parallel on the right, combining those we get,

2.00 Ω 6.00 Ω

4.00 Ω 3.00 Ω 6.00 Ω

2.00 Ω 6.00 Ω

4.00 Ω 2.00 Ω

1.00 2.00 3.00 6.00SR = Ω+ Ω+ Ω = Ω

1 1 13.00 6.00

2.00ppR

R⇒ == +Ω Ω

Ω

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Examples

20.58 – (cont.)Now, we have a pair of resistors in series, combining,

We now combine the two resistors in parallel,

We can now combine the final resistors which are in series,

2.00 Ω

4.00 Ω8.00 Ω 6.00 2.00 8.00SR = Ω+ Ω = Ω

2.00 Ω

2.67 Ω

1 1 14.00 8.

2.66700p

pRR Ω⇒ == +

ΩΩ

2.00 2.667 4.67SR = Ω+ Ω = Ω

Examples

20.66 – Three identical resistors are connected in parallel. The equivalent resistance increases by 700Ω when one resistor is removed and connected in series with the remaining two, which are still in parallel. Find the resistance of each resistor.Let us define the resistance in parallel

Removing one resistor and placing it in series with the other two:

Substituting our equations for R1 and R2 into the above,

11

11 1 1 33

1R R R R

RR

R= + + == ⇒

2

2 1

1 32 2

7

1

0

1 2

0

1

p

R R R

R

R R R

R

RR⇒ = + =

= +

= + =

Ω

3 1 7700 7002 3 6

600

R R R

R

= + Ω ⇒ = Ω

= Ω

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Internal Resistance of a Battery

• One way a battery is rated is by its voltage between its terminals.

• However, when a battery is connected to a load and there is current flowing, there is some internal resistance to a battery.

• The net result is to reduce the voltage between the terminals.

• The resistance inside the battery is constant, but the voltage drop of the terminals depends on the load:

•The voltage across the terminals is equivalent to the voltage over resistor R.

V

R

‐+

( )intV I R R= + ( )intterm

RV VR R

=+

Terminal Voltage andInternal resistance

For internal resistance of 0.01Ω

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Terminal Voltage andCurrent Through the Battery

• The more current through the battery, the lower the terminal voltage.

• If a 12V battery has an internal resistance of 0.010Ω and a current of 10A flowing through the battery. The terminal drop is

• If the current is 100A, then

( )int 12V 10A 0.0111.9V

term

term

V V IRV

= − = − Ω

=

( )int 12V 100 A 0.0111.0V

term

term

V V IRV

= − = − Ω

=

Kirchhoff’s Rules

• Sometimes circuits are too complex to analyze by reducing parallel and series resistors.

• For this we use two concepts:

• 1. Conservation of charge implies conservation of current.

• Let us assume that the currentflows in the circuit as drawn.

For conservation of current, the current flowing into a junction must equal the current flowing out.

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Kirchhoff’s Rules

• Take junction A. The current in must equal the current out.

• Label each branch of current.

• Find the currents in an out of each point:Point A:Point B:

• Note that both equations in this case are equivalent.

A

B

For conservation of current, the current flowing into a junction must equal the current flowing out.

Ileft I middle

Iright

right middle leftI I I= +

middle left rightI I I+ =

Kirchhoff’s Rules

• 2. Next we consider loops.

• Here there are two loops.

• Just like walking up and down hills, if you go in a loop the place you start/end must be at the same elevation.

• So, if you make a loop in a circuit, you come back to the same voltage (potential).

A

B

This is Kirchhoff’s Loop rule. The sum of potential rises and drops around any loop must be zero.

Ileft I middle

Iright

right middle leftI I I= +

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Kirchhoff’s Rules

• Go around each loop noting the potential rises and falls.

• Left loop: Let’s start at point A and go counter‐clockwise:

• Right loop, again starting at A, but let’s go clockwise:

• If we went counter‐clockwise, we would get the same answer!

A

B

This is Kirchhoff’s Loop rule. The sum of potential rises and drops around any loop must be zero.

Ileft I middle

Iright

right middle leftI I I= +

( ) ( )5V 2 3 0left leftI I− Ω − Ω =

( )( ) ( )4 12V 0rightI− − Ω + − =

( )12V 4 0rightI− Ω =

Kirchhoff’s Rules

• We can now find all the currents as we have three equations and three unknowns.

• This is an especially easy case as there is only one unknown in each loop equation.

A

B

Iright = 3.0AIleft = 1.0AImid = 2.0A

Because the current is positive, it moves in the direction assumed.

Ileft I middle

Iright

right middle leftI I I= +( ) ( )5V 3 2 0left leftI I− Ω − Ω =

( )12 V 4 0rightI− Ω =

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Examples20.73 – Two batteries, each with an internal resistance of

0.015Ω, are connected as in the drawing. In effect, the9.0‐V battery is being used to charge the 8.0‐V battery.What is the current in the circuit? For this we only need to apply Kirchhoff’s Loop Rule.Taking the current as flowing in the counter‐clockwisedirection and starting with the upper right corner:

The negative sign means that our original assumption about the current direction is wrong. So we have 33A going clockwise.

( ) ( )8.0V 0.015 0.015 9.00V 01.00 0.030 0

33A

I II

I

− Ω − Ω − =

− − == −

Examples

20.78 – Find the current in the 4.00Ω resistor in the drawing.Specify the direction of the current. We have three loops (one of which is the sum of theother two), and three branches with two junctions (oneof which is redundent). Let us label the currents as in thelower drawing. Thus the junction rule at B gives,

The loop rules for the left loop starting a D and goingclockwise gives,

The loop rules for the left loop starting a E and goingclockwise gives,

B

C

6.00 V

A

D

E

F

2.00 Ω

I1 I2

8.00 Ω

I3 4.00 Ω

9.00 V3.00 V+ + +

+ +

+

1 3 2I I I+ =

( ) ( )( )1 33V 2.00 6V 4.00 0I I− Ω + − − Ω =

( ) ( )3 29V 4.00 6 V 8.00 0I I− − Ω − − Ω =

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Examples

20.78 – (cont.) (1)

(2)

(3)We have three equations and three unknowns.We are only required to solve for I3, so, substituting (1) into (3),

Multiply equation (2) by four, , then adding to the above eqn.,

The negative sign indicates that the current is in the opposite direction from our initial assumption.

B

C

6.00 V

A

D

E

F

2.00 Ω

I1 I2

8.00 Ω

I3 4.00 Ω

9.00 V3.00 V+ + +

+ +

+

1 33 2 6 4 0I I− + + =

3 29 4 6 8 0I I− − − − =

1 3 2I I I+ =

( )3 1 3 139 4 6 8 15 12 80 0I I I I I− − − − + = ⇒ + + =

3

3

51 28 0

1.82A

I

I

+ =

= −

1 336 8 16 0I I− + =

Examples

20.104 – For the circuit shown in the drawing, find the current Ithrough the 2.00Ω resistor and the voltage V of the battery tothe left of this resistor. Let us define a current I1 flowing through the middle wire asshown. The junction rule gives,

The top loop gives (going counter‐clockwise),

Substituting into the loop rule equation we get

I1

1 3.00 AI I+ =

( ) ( )( ) ( )( )1

1

1

24V 6.00 3.00 A 4.00 3.00A 8.00 024 6 36 0

2.00 A

II

I

− Ω − Ω − Ω =

− − == −

5.00AI =

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Wheatstone Bridge

• Used to find resistances precisely.

• Rx is unknown but R1, R2 and R3 are accurately known.

• R3 is a variable resistor.

• Method is to vary R3 until the galvanometer does not deflect (i.e. points B and D are at the same voltage.

• At that time, the voltage drops across R3and R1 are equal: I1 R1= I3 R3

V

R1

R3

R2

Rx

A

B

D

CG

Wheatstone Bridge

• I1 R1= I3 R3

• So, it must be true that I2 R2= Ix Rx

• Since there is no voltage between B and D, I1= I2 and I3= Ix.

• I1R2= I3Rx• Let’s divide the two equations.

• Or,

V

R1

R3

R2

Rx

A

B

D

CG

3 3 31 1 1

1 2 3 2x x

I R RI R RI R I R R R

= ⇒ =

23

1xRR RR

=

R3 is very sensitive, so the galvanometer switch is closed only briefly to check if the current is zero.

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How does theGalvanometer Work?

• A DC Galvanometer measures current by using a coil spring, magnets and a loop of wire (more when we reach chapter 21).

• The rotation of the loop of wire gives the current passing through the system on some calibrated scale.

• The full scale current sensitivity of a galvanometer is the current required to move the needle (i.e. coil) deflect full scale.

• Depending on how the galvanometer is configured with resistors determines whether it reads current or voltage.

Measuring Current ‐ Ammeter

• Galvanometers can measure small currents. If the sensitivity is Im, it can read currents from about 0.02 Im up to Im. (e.g. if Im= 50 μA then the scale is 1 μA to 50 μA).

• Galvanometers (like batteries) have small internal resistances.

• If we want to measure current on a larger scale, a galvanometer is placed is parallel with a shunt resistor.

• Since the voltage drop across the galvanometer and resistor are the same:

G

R

R G GI R I R=

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Measuring Current ‐ Ammeter

• So,

• If the resistance of the galvanometer is known, then we can set R to be whatever resistance needed to make IR the scale of the ammeter.

• For example, if we want to measure up to 1 A, but the galvanometer can only take 50 μA and has an internal resistance of 30Ω, then we need IR to be 0.999950 A.

G

R

R G GI R I R=G G

R

I RRI

=

( )( )53

5.0 10 A 301.5 10

0.999950AR

−−

× Ω= = × Ω

Measuring Voltage ‐ Voltmeter

• A voltmeter also consists of a galvanometer and a resistor (usually large), but in series.

• Imagine we want to measure a full scale of V, then we would have

• For example, if we want the full scale to be 15V (again let’s take Im= 50 μA and RG= 30Ω.

• The internal resistance of the galvanometer is negligible.

GR

( )m G

Gm

V I R RVR RI

= +

= −

5

15V 30 300 k5 10 A

R −= − Ω = Ω×

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Examples20.84 – Two scales on a voltmeter measure voltages up to 20.0 and 30.0 V,

respectively. The resistance connected in series with the galvanometer is 1680Ωfor the 20.0‐V scale and 2930Ω for the 30.0‐V scale. Determine the coil resistance and the full‐scale current of the galvanometer that is used in the voltmeter.Using Ohm’s Law,

Subtracting one equation from the other,

Solving for Im,

Substituting back into either equation yields RG = 820Ω

( ) ( )1 1 2 2m G m GV I R R V I R R= + = +

2 1 1 2m mV V I R I R− = −

2 1

1 2

30V 20 V2930 1680

0.008A

m

m

V VIR R

I

− −= =

− Ω− Ω=

Examples20.107 – A galvanometer with a coil resistance of 12.0Ω and a full‐scale current of

0.150 mA is used with a shunt resistor to make an ammeter. The ammeter registers a maximum current of 4.00 mA. Find the equivalent resistance of the ammeter.Remember that the net current, is the current through the ammeter and the galvanometer:

The voltage across the galvanometer and shunt resistor is the same,

The equivalent resistance is that of the shunt resistor and galvanometer coil in parallel:

( )3

3

0.150 10 A 123.85 10 A

0.468

s s G G

Gs G

s

s

I R I R

IR RI

R

−

−

=

×= = Ω

×= Ω

3.85mAs GG sI II II I ⇒ = −+ ==

1 1 0.4501

eq s GeqR R RR⇒+ == Ω

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Capacitors aselements in a Circuit

• A Capacitor can be an element in a circuit, like a battery or resistor. It is denoted as or

• Since, q = CV, the voltage of a capacitor in a circuit depends on the capacitance and the charge:

• So, how does it work in a circuit?

qVC

=

Capacitors in Paralleland in Series

• Just as for resistors, when capacitors are in parallel, they see the same voltage, so thecapacitors will charge:

• So, we get,

• For capacitors in parallel:

1 2 3 1 2 3q q q CV C V C V+ + = + +

( )1 2 3tot Pq C V C C C V= = + +

( )1 2 3PC C C C= + +

C1 C2 C3V

CpV

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Capacitors in Paralleland in Series

• Just as for resistors, when capacitors are in series, they sum of the voltage drops must equalthe net voltage of the battery. So thecapacitors will charge:

• Since, the wire in between the capacitors begins as neutral, the charge must split evenly, so all capacitors are charged to the same level.

1 2 3V V V V= + +

1 2 3

1 1 1 1

totC C C C= + +

1 2 3S

q q q qC C C C

= + +

C1 C2 C3V

CSV

Capacitors in Paralleland in Series

• The rules for adding capacitors and for adding resistors are reversed:

Resistors Capacitors

Series

Parallel

1 2 3

1 1 1 1

SC C C C= + + +

1 2 3

1 1 1 1

PR R R R= + + +

1 2 3SR R R R= + + +

1 2 3SC C C C= + + +

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Charging and DischargingCapacitors in a Circuit

• If a Direct Current circuit contains a battery, resistor, an initially uncharged capacitor and a switch, then the capacitor will take time to charge:

• Where e is Euler’s constant (e=2.71828…)

• q0=CV0, where V0 is the voltage of the battery.

• The capacitor will reach full charge in infinite time (although in reality we can set a percentage of full charge to get an acceptable finite time).

( )/0 1 t RC

capq q e−= −

Charging and DischargingCapacitors in a Circuit

• If a Direct Current circuit contains a resistor, an initially charged capacitor and a switch, then the capacitor will take time to discharge:

• Where e is Euler’s constant (e=2.71828…)

• q0 is the initial charge on the battery.

• The capacitor will discharge in infinite time (although in reality we can set a low percentage of full charge to get an acceptable finite time).

/0

t RCcapq q e−=

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Charging and DischargingCapacitors in a Circuit

• Charging Capacitor

• Discharging Capacitor

• τ =RC, is the time constant of the circuit(s).

• When RC is small, the capacitor charges/discharges quickly

• When RC is large, the capacitor will charge/discharge slowly.

/0

t RCcapq q e−=

( )/0 1 t RC

capq q e−= −

Examples

20.97 – The circuit in the drawing contains two resistors andtwo capacitors that are connected to a battery via a switch.When the switch is closed, the capacitors begin to chargeup. What is the time constant for the charging process?Find the equivalent resistance and capacitance:

The time constant is the product of equivalent resistance and equivalent capacitance

1 21 2

1 1 1

1 1 13.0 F 6.0 F4.0 2.0

9.0 F 1.333

SP

SP

S p

C C CR R R

CR k k

C R k

μ μ

μ

= + = +

= + = +Ω Ω

= = Ω

( )( )3 6

2

1.333 10 9.0 10 F

1.2 10 sp SR Cτ

τ

−

−

= = × Ω ×

= ×

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Examples

20.98 – How many time constants must elapse before a capacitor in a series RC circuit is charged to 80.0% of its equilibrium charge?Use the Charging equation:

We want qcap to be 0.8q0,

( )( )

( )

/0 0

/

0.8 1

0 ln 0.2

ln 0.2 1.6

.2

1

t

t

q q e

e t

t

τ

τ

ττ τ

−

− ⇒ − =

= − =

= −

=

( )/0 1 t RC

capq q e−= −