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Simple Circuits Extra Drill Problems for AP Physics B & C

By — McGraw-Hill Professional
Updated on Feb 14, 2011

Practice problems and tests cannot possibly cover every situation that you may be asked to understand in physics.  However, some categories of topics come up again and again, so much so that they might be worth some extra review.  And that's exactly what these lessons are for - to give you a focused, intensive review of a few of the most essential physics topics.

Extra drill on difficult but frequently tested topics are:

We call them "drills" for a reason.  They are designed to be skill-building exercises and as such, they stress repetition and technique.  Working through these exercises might remind you of playing scales if you're a musician or of running laps around the field if you're an athlete.  Not much fun, maybe a little tedious, but very helpufl in the long run.

The questions in each drill are all solved essentially the same way.  Don't just do one problem after the other.....rather, do a couple, check to see that your answers are right, and then half an hour or a few days later, do a few more, just to remind yourself of the techniques involved.

Below are simple circuits problems.

How to Do It

Think "series" and "parallel." The current through series resistors is the same, and the voltage across series resistors adds to the total voltage. The current through parallel resistors adds to the total current, and the voltage across parallel resistors is the same.

The Drill

For each circuit drawn below, find the current through and voltage across each resistor.

    Note: Assume each resistance and voltage value is precise to two significant figures.
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  2. The Drill

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The Answers (A Step-by-Step Solution to 2 Is below.)

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Step-by-Step Solution to #2:

Start by simplifying the combinations of resistors. The 8 kΩ and 10 kΩ resistors are in parallel. Their equivalent resistance is given by

which gives Req = 4.4 kΩ

Next, simplify these series resistors to their equivalent resistance of 6.4 kΩ

6.4 kΩ (i.e., 6400 Ω) is the total resistance of the entire circuit. Because we know the total voltage of the entire circuit to be 10 V, we can use Ohm's law to get the total current

(more commonly written as 1.6 mA).

Now look at the previous diagram. The same current of 1.6 mA must go out of the battery, into the 2 kΩ resistor, and into the 4.4 kΩ resistor. The voltage across each resistor can thus be determined by V = (1.6 mA)R for each resistor, giving 3.2 V across the 2 kΩ resistor and 6.8 V across the 4.4 kΩ resistor.

The 2 kΩ resistor is on the chart. However, the 4.4 kΩ resistor is the equivalent of two parallel resistors. Because voltage is the same for resistors in parallel, there are 6.8 V across each of the two parallel resistors in the original diagram. Fill that in the chart, and use Ohm's law to find the current through each:

    I8k = 6.8 V/8000 Ω = 0.9 mA.
    I10k = 6.8 V/10,000 Ω = 0.7 mA.
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