Energy Crisis – Chem 021
Experimental demonstration 3.
These experiments are about electricity. We have 8 setups, groups of 5 will be formed
around 8 tables. We will form simple circuits and measure the current and voltage
(electrical potential difference) and calculate the electrical energy used.
Equipment:
Battery
Leads (Cu wires)
Switch
Light bulb
Meter WARNING: YOU MAY DAMAGE THE METER IF NOT CORRECLY
HOOKED UP. BEFORE HOOKING UP THE BATTERY, ASK THE TA OR ME TO
CHECK THE SETUP!!
Safety precaution: since we use low voltage batteries, there is no risk. However, do not
repeat any of this experiment using the higher voltage electricity outlets, which operate
on 110V or higher!
The batteries provide direct current (DC) in a closed circuit as opposed to the alternating
current (AC) of the outlets. The principles for a closed circuit are the same: electricity can
only flow in a fully completed circuit of conductors.
Materials are either good conductors (e.g. Ag, Au, Cu, Al), or insulators (many materials
including plastics, wood, rubber, table salt, glass, etc. A whole group of materials that are
conductors but provide more resistance, are the semiconductors, such as Si, and Ge.
The electric circuit is completed when the conductors of electricity (these are actually the
electrons, that are so important in chemistry) move from the negative pole of the battery
through the wires to the positive pole. They do their work while passing though the
circuit. Current (I) is
I= charge/time [its unit is A(mpere)]
We will deal with 1/1000th of this unit, the mA (miliampere).
The electrical potential difference (U) is provided by the battery that drives the charges
thrugh the circuit. Its unit is the V(olt)
Electrical units of energy:
E = Potential difference * charge = U*q
P(ower) = E(nergy)/t(ime)  E=Pt
Unit: 1 W = 1 J/s
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Power = E/t = Potential difference * charge/time = Potential difference * current
Units: 1W=1V(olt) * 1A(mpere)= 1 VA
A typical household machine uses a few amperes, say 5 A. This corresponds to the power
of 110V * 5 A = 550 W = 0.55 kW.
Energy is often calculated from power of electricity used:
Energy = Potential difference * current * time
When this machine, say vacuum cleaner, operates for 2 hours, the amount of work done
is:
Work = Power * t = 0.55 kW * 2 h= 1.1 kWh
Reminder: can we calculate the ever so important quantity:
Efficiency = work/(invested energy) ?
In the actual experiment try to establish the following:
The circuit can be broken with the switch.
The test is that the lamp should light up.
The meter should be correctly hooked up to measure the potential difference (a few
volts), and to measure the current (much less than 1A, but more than 1mA).
Calculate the electrical energy used. What data do you need in addition to U and I? Can
you estimate the efficiency of the light bulb?
List all the energy conversion steps, how many forms of energy are involved in this
experiment?
© Copyright, 2004, Miklos Kertesz
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