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Electric circuits, Current, and resistance - Physics Department ...

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<strong>Electric</strong> <strong>circuits</strong>, <strong>Current</strong>, <strong>and</strong> <strong>resistance</strong><br />

(Chapter 22 <strong>and</strong> 23)<br />

Acknowledgements: Several Images <strong>and</strong> excerpts are taken from<br />

College <strong>Physics</strong>: A strategic approach, Pearson Education Inc<br />

Dr. Mangala Singh, 1P22/92 Brock University


Dr. Mangala Singh, 1P22/92 Brock University


<strong>Current</strong><br />

If electric charge (e.g.<br />

electron) moves, we will say<br />

an electric current, I, is set<br />

to exist.<br />

I<br />

+<br />

A<br />

An electric current, I, is the<br />

rate at which net charge (∆Q)<br />

flows through a surface area A<br />

I<br />

∆Q<br />

= ∆ t<br />

<strong>Current</strong>’s units = C/S <strong>and</strong><br />

often written as A (Ampere)<br />

Electrons are actual current carriers in<br />

metals. They flow opposite to the direction<br />

of electric field or current<br />

Although current carriers (i.e., charges)<br />

could be “+” or “-”, the direction of<br />

current will be in the direction of a “+”<br />

charge flow (i.e., clockwise)<br />

Dr. Mangala Singh, 1P22/92 Brock University


<strong>Current</strong> <strong>and</strong> <strong>Current</strong> density<br />

n = number of mobile<br />

charges per unit volume<br />

q = charge on each current<br />

carriers<br />

∆x = distance they move in<br />

a time interval ∆t<br />

v d = speed (called drift<br />

velocity )<br />

J = current density<br />

Drift velocity v d ?<br />

I<br />

∆ x = v ∆t<br />

d<br />

+<br />

∆ Q = ( nA∆x)<br />

q<br />

= ( nAv ∆t)<br />

q<br />

I<br />

J<br />

d<br />

∆Q<br />

= = nqvd<br />

A<br />

∆t<br />

I<br />

= = nqvd<br />

A<br />

A<br />

Dr. Mangala Singh, 1P22/92 Brock University


Multiple choice questions<br />

1. The charge carriers in metals are<br />

A. electrons.<br />

B. positrons.<br />

C. protons.<br />

D. a mix of protons <strong>and</strong> electrons.<br />

2. A battery is connected to a resistor. Increasing the <strong>resistance</strong> of the resistor<br />

will<br />

A. increase the current in the circuit.<br />

B. decrease the current in the circuit.<br />

C. not affect the current in the circuit.<br />

3. A battery is connected to a resistor. As charge flows, the chemical<br />

energy of the battery is dissipated as<br />

A. current.<br />

B. voltage.<br />

C. charge.<br />

D. thermal energy.<br />

Dr. Mangala Singh, 1P22/92 Brock University


Problem: The discharge of the electric eel can transfer a charge of 2.0 mC in a<br />

time of 2.0 ms. What current, in A, does this correspond to?<br />

Solution:<br />

Step 1: To solve a problem, note each <strong>and</strong> every quantities mentioned in<br />

your problem. In this problem<br />

Q = 2.0 x10 -3 C<br />

Time = 2.0 x10 -3 s<br />

<strong>Current</strong> = ?<br />

Step 2: Identify relationship between these quantities<br />

Step 3: Rearrange your equation to find out the unknown quantity<br />

Add each <strong>and</strong> every numbers <strong>and</strong> compute.<br />

Warning: Calculator does what you ask for….If you insert wrong numbers or in<br />

a wrong manner, you will get a wrong answer!!! You must learn how to use your<br />

calculator properly<br />

I = Q/t = 2.0x10 -3 C/2.0x10 -3 s = 1A<br />

Dr. Mangala Singh, 1P22/92 Brock University


Batteries<br />

The potential difference between the<br />

terminals of a battery, often called<br />

the terminal voltage is often called<br />

battery’s electromotive force (emf)<br />

Wchem<br />

∆ Vbat<br />

= =<br />

q<br />

ξ<br />

Dr. Mangala Singh, 1P22/92 Brock University


Simple Circuits<br />

The current is determined by the<br />

potential difference <strong>and</strong> the <strong>resistance</strong> of<br />

the wire:<br />

I<br />

=<br />

∆<br />

V wire<br />

R<br />

battery<br />

Dr. Mangala Singh, 1P22/92 Brock University


Ohm’s law <strong>and</strong> <strong>resistance</strong><br />

A current density J & the electric field E are<br />

established in a conductor when a potential<br />

difference (V b -V a ) is maintained across the<br />

conductor<br />

J = σE ------- Ohm’s Law<br />

σ is called electrical conductivity of the materials<br />

Ohmic materials: In these materials the ratio of J<br />

<strong>and</strong> E is a constant “σ”. In other words, it is J<br />

varies linearly with E<br />

Value of conductivity of the materials are different for<br />

different materials<br />

V = V −V<br />

b<br />

V<br />

E = ⇒ V = El<br />

l<br />

V<br />

J = σ E = σ<br />

l<br />

l<br />

V = I ⇒ V = RI<br />

σ A<br />

l<br />

R =<br />

σ A<br />

a<br />

ρL<br />

R =<br />

A<br />

Dr. Mangala Singh, 1P22/92 Brock University


Resistance & Resistivity<br />

The <strong>resistance</strong> of a wire depends on its dimensions<br />

(length L & Cross sectional area A) <strong>and</strong> the resistivity of<br />

its material.<br />

R =<br />

ρL<br />

A<br />

Every material has a characteristic resistivity that only<br />

depends on the properties <strong>and</strong> composition of the<br />

material.<br />

Value of resistivity of the materials are different for<br />

different materials<br />

resistivity of the metal increase as temperature increase<br />

resistivity of semiconductors <strong>and</strong> insulators decreases<br />

as temperature increases<br />

Dr. Mangala Singh, 1P22/92 Brock University


Multiple choice<br />

A battery is connected to a wire, <strong>and</strong> makes a current in the wire. Which of the following changes<br />

would increase the current?<br />

(1) Increasing the length of the wire; (2) keeping the wire the same length, but making it thicker; (3)<br />

using a battery with a higher-rated voltage; (4) making the wire into a coil, but keeping its dimensions<br />

the same; (5) increasing the temperature of the wire.<br />

Problem 1:<br />

A. All of the above<br />

B. 1 <strong>and</strong> 5<br />

C. 1, 4, <strong>and</strong> 5<br />

D. 2 <strong>and</strong> 3<br />

E. None of the above<br />

A wire has <strong>resistance</strong> of 21 ohm. It is melted down <strong>and</strong> from the same volume of metal a new wire is<br />

made, that is three times longer than the original one. What is the <strong>resistance</strong> of new wire<br />

Problem 2:<br />

The filament of a 100-W bulb carries a current of 0.83 A at the normal operating voltage of 120 V.<br />

A. What is the <strong>resistance</strong> of the filament?<br />

B. If the filament is made of tungsten wire of diameter 0.035 mm, how long is the filament?<br />

Dr. Mangala Singh, 1P22/92 Brock University


Power in Circuits<br />

Dr. Mangala Singh, 1P22/92 Brock University


Energy <strong>and</strong> Power in Resistors<br />

Dr. Mangala Singh, 1P22/92 Brock University


Multiple choice<br />

A resistor is connected to a 3.0 V battery; the power dissipated in the resistor is 1.0 W. The battery is<br />

traded for a 6.0 V battery. The power dissipated by the resistor is now<br />

A. 1.0 W<br />

B. 2.0 W<br />

C. 3.0 W<br />

D. 4.0 W<br />

Problem 1: An electric blanket has a wire that runs through the interior. A current causes energy to be<br />

dissipated in the wire, warming the blanket. A new, low-voltage electric blanket is rated to be used at 18<br />

V. It dissipates a power of 82 W. What is the <strong>resistance</strong> of the wire that runs through the blanket?<br />

Dr. Mangala Singh, 1P22/92 Brock University


Example Problem<br />

An electric blanket has a wire that runs through the interior. A current causes energy to be dissipated in<br />

the wire, warming the blanket. A new, low-voltage electric blanket is rated to be used at 18 V. It<br />

dissipates a power of 82 W. What is the <strong>resistance</strong> of the wire that runs through the blanket?<br />

Summary<br />

Dr. Mangala Singh, 1P22/92 Brock University


Conservation of <strong>Current</strong><br />

∑<br />

I<br />

in<br />

= ∑<br />

I<br />

out<br />

Dr. Mangala Singh, 1P22/92 Brock University


Multiple choice questions<br />

1. Rank the bulbs in the following circuit according to their brightness, from brightest to dimmest.<br />

A. A > B = C > D<br />

B. A = B = C = D<br />

C. A = D > B = C<br />

D. B = C > A > D<br />

Dr. Mangala Singh, 1P22/92 Brock University


Drawing Circuit Diagrams<br />

Dr. Mangala Singh, 1P22/92 Brock University


Kirchhoff’s Law<br />

∑<br />

I<br />

in<br />

= ∑<br />

I<br />

out<br />

∆ V = ∑ V =<br />

loop<br />

i<br />

0<br />

Dr. Mangala Singh, 1P22/92 Brock University


Using Kirchhoff’s Laws<br />

Dr. Mangala Singh, 1P22/92 Brock University


Multiple choice questions<br />

1. The diagram below shows a segment of a circuit. What is the<br />

current in the 200Ω resistor?<br />

A. 0.5 A<br />

B. 1.0 A<br />

C. 1.5 A<br />

D. 2.0 A<br />

E. There is not enough information to decide.<br />

2. The diagram below shows a circuit with two batteries <strong>and</strong> three resistors. What is the potential<br />

difference across the 200Ω resistor?<br />

A. 2.0 V<br />

B. 3.0 V<br />

C. 4.5 V<br />

D. 7.5 V<br />

E. There is not enough information to decide.<br />

Dr. Mangala Singh, 1P22/92 Brock University


Series Resistors<br />

Dr. Mangala Singh, 1P22/92 Brock University


Parallel Resistors<br />

Dr. Mangala Singh, 1P22/92 Brock University


Problems<br />

1. There is a current of 1.0 A in the circuit below. What is the<br />

<strong>resistance</strong> of the unknown circuit element?<br />

2. What is the current out of the battery?<br />

Dr. Mangala Singh, 1P22/92 Brock University


Analyzing Complex Circuits<br />

Dr. Mangala Singh, 1P22/92 Brock University


Problems<br />

1. In the circuit shown below:<br />

A. Rank in order, from most to least bright, the brightness of<br />

bulbs A–D. Explain.<br />

B. Describe what, if anything, happens to the brightness of bulbs<br />

A, B, <strong>and</strong> D if bulb C is removed from its socket. Explain.<br />

Dr. Mangala Singh, 1P22/92 Brock University


Problems<br />

In the circuit shown below:<br />

A. How much power is dissipated by the 12 Ω resistor?<br />

B. What is the value of the potential at points a, b, c, <strong>and</strong> d?<br />

Dr. Mangala Singh, 1P22/92 Brock University


Battery <strong>and</strong> its internal <strong>resistance</strong> (r)<br />

As the charges passes through<br />

from – to + terminal of the<br />

battery its potential energy<br />

increases<br />

The current in the circuit<br />

-Depends both external<br />

<strong>resistance</strong> to the battery<br />

-And its internal <strong>resistance</strong><br />

I<br />

=<br />

ξ<br />

R + r<br />

Due to internal <strong>resistance</strong> of the<br />

battery, its potential decreases<br />

by Ir, wherein I is the current.<br />

Terminal voltage of the battery<br />

= V +<br />

-V -<br />

= ξ-Ir<br />

Note: Terminal voltage may exceed the emf<br />

by an amount Ir – when the current is<br />

opposite the emf as in the case of charging a<br />

battery with another source of emf.<br />

Power dissipated, P, in the load<br />

<strong>resistance</strong> R<br />

ξ<br />

( R + r)<br />

2<br />

P 2<br />

= I R = R 2<br />

Dr. Mangala Singh, 1P22/92 Brock University


Problems:<br />

1. Show that maximum power lost in the load resistor R occurs<br />

when R =r, i.e., when the value of load <strong>resistance</strong> matches the<br />

internal <strong>resistance</strong> of the battery.<br />

Dr. Mangala Singh, 1P22/92 Brock University

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