Ohm’s Law Practice Worksheet With Answers
Suppose you were to build this circuit and take measurements of current through the resistor and voltage across the resistor:
Recording these numerical values in a table, the results look something like this:
|0.22 A||0.66 V|
|0.47 A||1.42 V|
|0.85 A||2.54 V|
|1.05 A||3.16 V|
|1.50 A||4.51 V|
|1.80 A||5.41 V|
|2.00 A||5.99 V|
|2.51 A||7.49 V|
Plot these figures on the following graph:
What mathematical relationship do you see between voltage and current in this simple circuit?
Plot the relationships between voltage and current for resistors of three different values (1 Ω, 2 Ω, and 3 Ω), all on the same graph:
What pattern do you see represented by your three plots? What relationship is there between the amount of resistance and the nature of the voltage/current function as it appears on the graph?
Advanced question: in calculus, the instantaneous rate-of-change of an (x,y) function is expressed through the use of the derivative notation: [dy/dx]. How would the derivative for each of these three plots be properly expressed using calculus notation? Explain how the derivatives of these functions relate to real electrical quantities.
One style of light bulb, very different from the “incandescent” design which works on the principle of a super-heated wire filament emitting light, is called a gas discharge tube. In this design of light bulb, light is produced by the direct “excitation” of gas molecules as electric current passes between two electrodes:
Both types of light bulbs have interesting voltage/current plots, neither one being identical to the voltage/current plot of a resistor. First, the voltage/current plot for an incandescent light bulb:
Next, the voltage/current plot for a gas-discharge light bulb:
Based on these two graphs, what can you say about the electrical resistance of each bulb type over its operating range?
There are two basic Ohm’s Law equations: one relating voltage, current, and resistance; and the other relating voltage, current, and power (the latter equation is sometimes known as Joule’s Law rather than Ohm’s Law):
In electronics textbooks and reference books, you will find twelve different variations of these two equations, one solving for each variable in terms of a unique pair of two other variables. However, you need not memorize all twelve equations if you have the ability to algebraically manipulate the two simple equations shown above.
Demonstrate how algebra is used to derive the ten “other” forms of the two Ohm’s Law / Joule’s Law equations shown here.
The brightness of a light bulb - or the power dissipated by any electrical load, for that matter - may be varied by inserting a variable resistance in the circuit, like this:
This method of electrical power control is not without its disadvantages, though. Consider an example where the circuit current is 5 amps, the variable resistance is 2 Ω, and the lamp drops 20 volts of voltage across its terminals. Calculate the power dissipated by the lamp, the power dissipated by the variable resistance, and the total power provided by the voltage source. Then, explain why this method of power control is not ideal.
A modern method of electrical power control involves inserting a fast-operating switch in-line with an electrical load, to switch power on and off to it very rapidly over time. Usually, a solid-state device such as a transistor is used:
This circuit has been greatly simplified from that of a real, pulse-control power circuit. Just the transistor is shown (and not the “pulse” circuit which is needed to command it to turn on and off) for simplicity. All you need to be aware of is the fact that the transistor operates like a simple, single-pole single-throw (SPST) switch, except that it is controlled by an electrical current rather than by a mechanical force, and that it is able to switch on and off millions of times per second without wear or fatigue.
If the transistor is pulsed on and off fast enough, power to the light bulb may be varied as smoothly as if controlled by a variable resistor. However, there is very little energy wasted when using a fast-switching transistor to control electrical power, unlike when a variable resistance is used for the same task. This mode of electrical power control is commonly referred to as Pulse-Width Modulation, or PWM.
Explain why PWM power control is much more efficient than controlling load power by using a series resistance.
What would happen if a wire having no resistance at all (0 Ω) were connected directly across the terminals of a 6-volt battery? How much current would result, according to Ohm’s Law?
Suppose we were to short-circuit a 6-volt battery in the manner just described and measure 8 amps of current. Why don’t the calculated figures from the previous paragraph agree with the actual measurement?
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