Essential Physics

Solving compound circuit problems


More complicated circuits	How do you determine the equivalent resistance for circuits that have more than two resistances? The answer is that you break the circuit apart into individual series or parallel circuits and solve them individually. Then combine larger and larger elements of the circuit as series and parallel circuits.

An example	How do you calculate the equivalent resistance of the three-resistor circuit above? Start by looking for a small portion of the circuit that, by itself, looks like a series or parallel circuit. In this example, the 10 Ω and 15 Ω resistors are in parallel, so we use equation (17.4) to determine that their equivalent resistance is 6 Ω. We replace the 10 Ω and 15 Ω resistors with a single 6 Ω resistor and look at the circuit again. Now we see that this 6 Ω resistor is in series with the 20 Ω resistor, so we use equation (17.3) to determine that their equivalent resistance is 26 Ω. Since we have now reduced the three-resistor circuit to a single, equivalent resistor, this is the answer.

A more complicated example	The circuit above contains four resistors but we still use the same method for finding its equivalent resistance. In this case, the 5 Ω and 10 Ω resistors are in series so they add to 15 Ω using equation (17.3). This 15 Ω resistance is in parallel with the 20 Ω resistor, so we use equation (17.4) to calculate their equivalent resistance of 8.6 Ω. Finally, this 8.6 Ω resistance is in series with the 30 Ω resistor, which combine to make 38.6 Ω. It may take a few steps to calculate the equivalent resistance of 38.6 Ω, but all we needed were the equations for series and parallel resistance!
Test your knowledge	Suppose that you have a battery, connecting wires, and three identical light bulbs. Each bulb behaves like an ideal resistor (that is, obeys Ohm’s law). Which of the circuits shown here will produce the most light in total?
	Circuit A will produce the most light in total. Carefully trace the routes that positive current can follow once leaving the battery’s positive terminal. In Circuit A, current can flow through any of the three bulbs and immediately return to the battery. This means that the bulbs are in parallel. In Circuit B, by contrast, current must flow through the uppermost bulb to reach the middle one; and it must flow through the middle bulb to reach the lowest one. These bulbs are therefore in series. Circuit C is an in-between case, with two bulbs in parallel and the third in series with them. Power is maximized when the bulbs are in parallel, and brightness increases when power does. Therefore Circuit A will produce the most light.


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