Current
Current

Current is defined as the flow of positive charge.

I = Q/t
I: current in Amperes or Amps (A)
 Q: charge in Coulombs (C)
 t: time in seconds (s)

Electron Flow

In a normal electrical circuit, it is the electrons
that carry the charge.

So if the electrons one way, which way does the
current move?
Problem

How many electrons per hour flow past a point in a circuit if it
bears 11.4 mA of direct current?
EMF

Cells convert chemical energy into electrical energy.

The potential difference (voltage) provided by a cell is
called its electromotive force (or emf).

The emf of a cell is constant, until near the end of the
cell’s useful lifetime.

Misnomer: The emf is not really a force.
Battery

A battery is composed of more than one cell in
series.

The emf of a battery is the sum of the emf’s of
the cells.
Circuit Components
Voltmeter
Ammeter
Switch
Ohmmeter
Problem


If a typical AA cell has an emf of 1.5 V, how much emf do 4 AA
cells provide?
Draw the battery composed of these 4 cells.
Problem


Draw a single loop circuit that contains a cell, a light bulb, and a
switch.
Put a voltmeter in the circuit so it reads the potential difference
across the light bulb.
Series

Series components are put together so that all
the current must go through each one

Bulbs in series all have the same current.
Parallel

Parallel components are put together so that the
current divides, and each component gets only a
fraction of it.

Bulbs in parallel do not have the same current.
Their current is proportional to the resistance of
the paths.
Problem

Draw a circuit with a cell and two bulbs in series.
Problem

Draw a circuit having a cell and four bulbs. Exactly two of the
bulbs must be in parallel.
Mini Lab A

Draw a circuit containing one cell, one bulb, and
a switch. Create this with the available
components. Measure the voltage across the cell
and then across the bulb. What do you observe?
Mini Lab B

Draw a circuit containing a cells and two bulbs
in series. Create this with the available
components.

What do you observe happens to the other
bulb’s brightness when you unscrew one of
them?

Measure the voltage across the battery and
across each bulb. What do you observe?
Mini Lab C

Draw a circuit containing a cells and two bulbs
in parallel. Create this with the available
components.

What do you observe happens to the bulbs
when you unscrew one bulb?

Measure the voltage across the battery and
across each bulb. What do you observe?
Mini Lab D

Draw a circuit containing a cells, three bulbs in
series, and a switch which shorts the last bulb
when it is closed. Create this with the available
components.

What do you observe happens to the bulbs
when you close the switch?
Ohm’s Law
Resistors

Resistors are devices put in circuits to reduce the
current flow.

Resistors are built to provide a measured
amount of “resistance” to electrical flow, and
thus reduce the current.
Circuit Components
Resistor
Resistance


Resistance depends on resistivity and on
geometry of the resistor.
R = L/A
 : resistivity (Ω m)
 L: length of resistor (m)
 A: cross sectional area of resistor (m2)


Unit of resistance: Ohms (Ω)
Problem

What is the resistivity of a substance which has a resistance of
1000 Ω if the length of the material is 4.0 cm and its cross
sectional area is 0.20 cm2?
iClicker
Two concentric circular loops of radii b and 2b, made of the same type of wire, lie
in the plane of the page, as shown above. The total resistance of the wire loop
of radius b is R. What is the resistance of the wire loop of radius 2b ?
(A) R/4
(B) R/2
(C) R
(D) 2R
(E) 4R
Problem

What is the resistance of a mile of copper wire if the diameter is
5.0 mm? (resistivity of copper is 1.72 x 10-8)
iClicker
The five resistors shown below have the lengths and cross-sectional
areas indicated and are made of material with the same resistivity.
Which resistor has the least resistance?
Ohm’s Law

Resistance in a component in a circuit causes
potential to drop according to the equation:

ΔV = IR
ΔV: potential drop (Volts)
 I: current (Amperes)
 R: resistance (Ohms)

Ohmmeter

Measures Resistance.

Placed across a resistor when no current is
flowing.
Ammeter

An ammeter measures current.

It is placed in the circuit in a series connection.

An ammeter has very low resistance, and does
not contribute significantly to the total
resistance of the circuit.
Power

P = W/t

P = ΔE/Δt

Units
Watts
 Joules/second

Power in Circuits

P = IΔV
P: power (W)
 I: current (A)
 ΔV: potential difference (V)


P = I2R

P = (ΔV)2/R
iClicker
The product 2 amperes x 2 volts x 2 seconds
is equal to
(A) 8 coulombs
(B) 8 newtons
(C) 8 joules
(D) 8 calories
(E) 8 newton-amperes
iClicker
The circuit shown above left is made up of a variable resistor and a battery with
negligible internal resistance. A graph of the power P dissipated in the resistor
as a function of the current I supplied by the battery is given above right.
What is the emf of the battery?
(A) 0.025 V
(B) 0.67 V
(C) 2.5 V
(D) 6.25 V
(E) 40 V
Problem

How much current flows through a 100-W light bulb connected
to a 120 V DC power supply? What is the resistance of the bulb?
iClicker
A certain coffeepot draws 4.0 A of current when it is
operated on 120 V household lines. If electrical energy
costs 10 cents per kilowatt-hour, how much does it cost
to operate the coffeepot for 2 hours?
(A) 2.4 cents
(B) 4.8 cents
(C) 8.0 cents
(D) 9.6 cents
(E) 16 cents
Resistors in Circuits

Resistors can be placed in circuits in a variety of
arrangements in order to control the current.

The overall resistance of a specific grouping of resistors
is referred to as the equivalent resistance.

The equivalent resistance in a circuit determine the
current output from the battery.
Resistors in Circuits

Arranging resistors in series increases the resistance and
causes the current to be reduced.

Arranging resistors in parallel reduces the resistance and
causes the current to increase.
Resistors in Series
Resistors in Parallel
iClicker
The electrical resistance of the part of the circuit shown between
point X and point Y is
(A) 4/3 
(B) 2 
(C) 2.75 
(D) 4 
(E) 6 
iClicker
Which two arrangements of resistors shown above have the same
resistance between the terminals?
(A) I and II
(B) I and IV
(C) II and III
(D) II and IV
(E) III and IV
iClicker
A lamp, a voltmeter V, an ammeter A, and a battery with zero internal resistance
are connected as shown above. Connecting another lamp in parallel with the
first lamp as shown by the dashed lines would
(A) increase the ammeter reading
(B) decrease the ammeter reading
(C) increase the voltmeter reading
(D) decrease the voltmeter reading
(E) produce no change in either meter reading
Kirchoff’s Rules
Problem

Draw a circuit containing, in order (1) a 1.5 V cell, (2) a
100-Ω resistor, (3) a 330-Ω resistor in parallel with a
100-Ω resistor (4) a 560-Ω resistor, and (5) a switch.

Calculate the equivalent resistance.

Calculate the current through the cell.

Calculate the current through the 330-Ω resistor.
iClicker
In the diagrams above, resistors R1 and R2 are shown in two different
connections to the same source of emf  that has no internal resistance. How
does the power dissipated by the resistors in these two cases compare?
(A) It is greater for the series connection.
(B) It is greater for the parallel connection.
(C) It is the same for both connections
(D) It is different for each connection, but one must know the values of R1 and
R2 to know which is greater
(E) It is different for each connection, but one must know the value of  to know
which is greater.
Junction Rule

Kirchoff’s 1st rule is also called the “junction rule”.

The sum of the currents entering a junction equals the
sum of the currents leaving the junction.

This rule is based upon conservation of charge.

Amount of current flowing in any direction is
proportional to the rersistance along the path
iClicker
When there is a steady current in the circuit, the amount of charge
passing a point per unit of time is
(A) the same everywhere in the circuit
(B) greater at point X than at point Y
(C) greater in the 1  resistor than in the 2  resistor
(D) greater in the 1  resistor than in the 3  resistor
(E) greater in the 2  resistor than in the 3  resistor
Loop Rule

Kirchoff’s 2nd rule is also referred to as the
“loop rule”.

The net change in electrical potential in going
around one complete loop in a circuit is equal to
zero.
iClicker
Kirchhoff’s loop rule for circuit analysis is an
expression of which of the following?
(A) Conservation of charge
(B) Conservation of energy
(C) Ampere's law
(D) Faraday's law
(E) Ohm's law
Ohm’s Law Graph

Make a table of current and resistance data and graph the data such that voltage is the
slope of a best-fit line.

Wire a circuit with a cell and one or more resistors.

Calculate and record the resistance.

Measure and record the corresponding current. Do this 5 times without duplicating
your resistance values.

Rearrange the equation ΔV = IR so that ΔV is the slope of a linear equation.

Construct a graph from your data that corresponds to this rearranged equation.

Calculate and clearly report the slope of the line. How does this compare to the emf of
1.5 V for a D-cell?
Terminal Voltage

When a current is drawn from a battery, the voltage
across its terminals drops below its rated EMF.

The chemical reactions in the battery cannot supply
charge fast enough to maintain the full EMF.

Thus the battery is said to have an internal resistance,
designated r.
Terminal Voltage and EMF

A real battery is then modeled as if it were a perfect
emf, ε. in series with a resistor r.

Terminal voltage Vab

When no current is drawn from the battery, the
terminal voltage equals the emf.

When a current I flows from the battery, there is an
internal drop in voltage equal to Ir, thus the terminal
voltage (actual voltage delivered) is Vab = ε - Ir
iClicker
In the circuit shown above, what is the value of the potential difference between
points X and Y if the 6-volt battery has no internal resistance?
(A) 1 V
(B) 2 V
(C) 3 V
( D) 4 V
(E) 6V
iClicker
An immersion heater of resistance R converts electrical
energy into thermal energy that is transferred to the
liquid in which the heater is immersed. If the current in
the heater is I, the thermal energy transferred to the
liquid in time t is
(A) Irt
(B) I2Rt
(C) IR2t
(D) IRt2
(E) IR/t