PHYSICS
ELECTRICITY Analyzing Simple DC Resistive Circuits
Ohm's Law Kirchhoff's Laws
V = IR Iin = Iout Vup = Vdown
A Simple Circuit
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Example Problem #1 Ohm's Law is all that is needed to solve this simple circuit. If any two pieces of the "puzzle" are known, substitution into Ohm's Law does the rest. For example: Suppose the battery voltage is 12 V and the resistance is 4 W. Then the current I = V/R = 12V/4W = 3 A. |
As the problems become more complex, the number of ways you can correctly solve the problem increases. It is just like solving a jigsaw puzzle. Some people prefer to collect puzzle pieces by color and start that way. Other people start by assembling the borders. Each way is a start, and as you continue you build more and more connections for assembling the rest of the puzzle.
Likewise for circuit problems, each part of the problem you solve gives you more ways for finding the rest. Solving more than one way allows you to check your work. But you must follow the correct rules for the type of problem that you are solving. The rules for series circuits and the rules for parallel circuits are not all the same. All circuits use Ohm's Law and Kirchhoff's Laws, but the rest of the rules are as unique and different as the rules for baseball and football. Try to keep them straight.
In each problem, some of the information is given, and others must be found. Remember the following:
Series Circuits [one path]
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Example #2 Suppose the battery voltage VBattery is 10 V, R1 is 5 W and R2 is 5 W. What is the current IT through the ammeter? What is V1 (the voltage across resistor R1), and I1 (the current through R1)? Also, what is V2 and I2?
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Solution: Perhaps the easiest way to start this problem is to solve for the total resistance.
RT = R1 + R2 = 5 W + 5 W = 10 W
Since VBattery = VT, We can then apply Ohm's Law to the entire circuit to get IT.
IT = VT/RT = 10 V/10 W = 1 A
Then, since the current is the same everywhere in a series circuit, I1 = I2 = IT = 1 A.
Ohm's Law can then be applied to the positions of both resistor 1 and 2 because at each position two of the three parts are known. This solves for V1 and V2.
V1 = I1R1 = (1 A)(5 W) = 5 V and V2 = I2R2 = (1 A)(5 W) = 5 V
Checking, using Kirchhoff's 2nd Law, V1 + V2 = 5 V + 5 V = 10 V = VT, as required. The problem is solved.
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Example #3 Suppose the battery voltage VBattery is 12 V, R1 is 2 W and R2 is 4 W. What is the current IT through the ammeter? What is V1 (the voltage across resistor R1), and I1 (the current through R1)? Also, what is V2 and I2?
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Solution: Again the easiest way to start this problem is to solve for the total resistance.
RT = R1 + R2 = 2 W + 4 W = 6 W
Since VBattery = VT, We can then apply Ohm's Law to the entire circuit to get IT.
IT = VT/RT = 12 V/6 W = 2 A
Then, since the current is the same everywhere in a series circuit, I1 = I2 = IT = 2 A.
Ohm's Law can then be applied to the positions of both resistor 1 and 2 because at each position two of the three parts are known. This solves for V1 and V2.
V1 = I1R1 = (2 A)(2 W) = 4 V and V2 = I2R2 = (2 A)(4 W) = 8 V
Checking, using Kirchhoff's 2nd Law, V1 + V2 = 4 V + 8 V = 12 V = VT, as required. The problem is solved.
Note that with the current equal to 2A, the voltage values and resistance values at each resistor are no longer the same as in Example #2. However, they are still in proportion to each other.
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Example #4 Suppose the battery voltage is 12 V, and the resistance of R1 = R2, but the values are unknown. Suppose the resistance R3 = 4 W, and the current through R3 is I3 = 2 A. Find R1, R2, I1, I2, I3, V1, V2, and V3. |
Solution: Any time two of the three parts of Ohm's Law are known for the entire circuit, or for any part of a circuit, the third part can be found. Since R3 and I3 are known, then V3 = I3R3 = (2 A)(4 W) = 8 V. That leaves only 12 V - 8 V = 4 V left to drop over the remaining two resistors. Since these two resistors are equal, and since voltage drops over resistors in series are proportional to the resistances, then the two voltage drops, V1 and V2, must be equal to each other. V1 = V2 and V1 + V2 = 4 implies that V1 = V2 = 2 V. Since the current in a series circuit is the same everywhere, then I1 = I2 = 2 A. Ohm's Law now can be used to calculate R1 and R2. R1 = V1/I1 = R2 = V2/I2 = 1 W.
Parallel Circuits [two or more paths]
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Example #5 The battery voltage is 10 V, and the resistance of R1= 3W while the resistance of R2 = 6W. Find R1, R2, I1, I2, IT, V1, V2, and RT.
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Solution: The easiest part of a parallel circuit problem is the voltage. Kirchhoff's law says that the voltage gain equals the voltage drop, and both resistors have the full 10 V drop. Each current electron will go only through one resistor. Therefore,
V1 = V2 = VBattery = 10 V
But the total current is not so easily solved for. You must first find the total resistance. For a parallel circuit it is conductance (1/R) that adds. Hence (remembering that adding fractions requires a common denominator):
1/RT = 1/R1 + 1/R2 = 1/3W + 1/6W = 2/6W + 1/6W = 3/6W = 1/2W, so 1/RT = 1/2W
Be careful here, since this is the reciprocal of the total parallel voltage. You must now flip over both sides to get
RT = 2W
But wait, does this mean that the total resistance is less than the resistance of either of the two resistors? Yes it does. The combined resistance is always less than that of the smallest resistor in parallel. Why? Because the second resistor adds a second path for current to follow, and the resistance of this path has to be less than that of no path at all!
Now we can find IT
IT = VT/RT = 10V/2W = 5 A
Of course, this is not the only way to find IT. In fact there is an easier way. We were given R1 and immediately wrote down V1. So, using Ohm's Law,
I1 = V1/R1 = 10V/3W = 10/3 A
Likewise, I2 = V2/R2 = 10V/6W = 5/3 A
and IT = I1 + I2 = 10/3 A + 5/3 A = 15/3 A = 5 A
Since both methods yield the same result for IT, we can safely assume that our work is correct and the problem is solved.
Combination Circuits
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Example #6 Suppose the battery voltage is 10 V, R1 and R2 are again in parallel to each other, but R3 has been added in series. This is a combination circuit. You must use the series rules for the series parts and the parallel rules for the parallel parts. As in the previous problem, let R1= 3 W and R2 = 6 W. Also, let R3 = 3 W. Find I1, I2, I3, V1, V2, and V3.
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Solution: The key to this problem is to realize that a portion of a circuit can be replaced with a simpler configuration with the same properties, in this case, the parallel resistors with a single resister with the equivalent resistance. By finding this equivalent resistance, we can solve the circuit as though it is two resistors in series, RParallel and R3. RParallel is found by remembering that for parallel circuits, it is conductance (the inverse of resistance) that adds.
Thus, 1/RParallel = 1/R1 + 1/R2 = 1/3W + 1/6W = 2/6W + 1/6W = 3/6W = 1/2W.
Flipping over the proportion, 1/RParallel = 1/2W, yields RParallel = 2W.
Thus, the entire parallel "element" can be replaced by a 2W resistor in series with R3.
Then RT = RParallel + R3 = 2W + 3W = 5W.
IT = VT/RT = 10V/5W = 2A, which is also equal to I3 because the current in a series circuit is the same everywhere.
Then V3 = I3R3 = (2A)(3W) = 6V,
and since the voltage drop across resistors in parallel is the same,
V1 = V2 = Vparallel =VT - V3 = 10V - 6V = 4V.
I1 = V1/R1 = 4V/3W = 4/3 A = 1.33A.
Using Kirchhoff's 1st Law: I2 = IT - I1 = 2/3 A = 0.67A.
All values have been found.
More Complex Circuits
Complex circuits are made out of simple parts. They are resolved by solving the smallest parts first, and then substituting their equivalent values in the next smallest part, until the entire problem is solved. The individual pieces are either in series or in parallel.
These problems are not more difficult, they just take longer. It is similar to an algebraic expression with nested parentheses; you solve the innermost parenthesis first and then move outward.
ELECTRICITY Analyzing Simple DC Resistive Circuits 05/09/08
