Basic Electricity

Ohm’s Law Practice Worksheet With Answers

23 questions By Tony R. Kuphaldt

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Question 22 of 23

What will happen to the brightness of the light bulb if the switch in this circuit is suddenly closed?

Reveal answer

Ideally, there will be no change whatsoever in the light bulb’s brightness when the switch is closed, because voltage sources are supposed to maintain constant voltage output regardless of loading. As you might have supposed, though, the additional current “drawn” by the resistor when the switch is closed might actually cause the lamp to dim slightly, due to the battery voltage “sagging” under the additional load. If the battery is well oversized for the application, though, the degree of voltage “sag” will be inconsequential.

Notes:
This question illustrates a disparity between the ideal conditions generally assumed for theoretical calculations, and those conditions encountered in real life. Truly, it is the purpose of a voltage source to maintain a constant output voltage regardless of load (current drawn from it), but in real life this is nearly impossible. Most voltage sources exhibit some degree of “sag” in their output over a range of load currents, some worse than others.

In this example, it is impossible to tell how much the voltage source’s output will “sag” when the switch is closed, because we have no idea of what the resistor’s current draw will be compared to that of the light bulb, or what the voltage source’s rated output current is. All we can say is that theoretically there will be no effect from closing the switch, but that in real life there will be some degree of dimming when the switch is closed.
Question 23 of 23

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?

Reveal answer

Ohm’s Law would suggest an infinite current (current = voltage divided by zero resistance). Yet, the experiment described yields only a modest amount of current.

If you think that the wire used in the experiment is not resistance-less (i.e. it does have resistance), and that this accounts for the disparity between the predicted and measured amounts of current, you are partially correct. Realistically, a small piece of wire such as that used in the experiment will have a few tenths of an ohm of resistance. However, if you re-calculate current with a wire resistance of 0.1 Ω, you will still find a large disparity between your prediction and the actual measured current in this short-circuit.

Follow-up question #1: explain why wire resistance alone does not explain the modest short-circuit current.

Follow-up question #2: identify at least one safety hazard associated with a real experiment such as this.

Notes:
Remind students that short-circuit testing of electrical power sources can be dangerous. A student of mine once stuffed a 6-volt “lantern” battery in his tool pouch, only to have it discharge smoke an hour later, after the battery terminals had been shorted together by a wrench handle!

No, Ohm’s Law is not being cheated here: shorting a voltage source with a 0 Ω conductor will not result in infinite current, because there are other sources of resistance in such a circuit. The task here is to determine where those sources might be, and how they could be located.