OK. These are actually AC circuits. Since the loads are almost purely resistive, i.e., there are no capacitances or inductances (or they are small enough to be negligible), and since the rms AC voltage and current behave in purely resistive circuits as DC voltage and current do, the two circuits shown above are equivalent to the corresponding DC circuits. The AC from the wall is sinusoidal. The rms (root-mean-square) voltage for a sinusoid is 0.707V

_{p}, where V_{p}is the peak voltage. Similarly, the rms current through a resistor is 0.707i_{p}, where i_{p}is the peak current. Theseeffectivevalues correspond to the DC values that would give the same power dissipation in the resistor. These are slightly different from theaveragevoltage and current, which are 0.639V_{p}and 0.639I_{p}for a sinusoid. For AC from the wall, the rms voltage is approximately 120 V, and the average voltage is about 110 V.Each board has three 40-watt bulbs, connected as shown by the resistor circuits painted on it. The board on the left has the bulbs arranged, of course, in parallel, and the board on the right has them in series. Since power, P, equals iV, P/V = i, so at 120 V, a 40-watt bulb draws 1/3 A. (The units in

iV are (C/s)(N-m/C), or J/s, which are watts.) For a given resistance, V = iR, so the bulb’s resistance (when it has 120 volts across it) is 120/(1/3), or 360 ohms. (We also know by the two equations above that P = i^{2}R, which gives R as 40/(1/9), or 360 ohms.)When the bulbs are connected in parallel, each bulb has 120 V across it, each draws 1/3 A, and each dissipates 40 watts. In this circuit, all bulbs glow at their full brightness. The total power dissipated in the circuit is three times 40, or 120 watts (or 3(1/3) A × 120 V = 120 W).

In the series circuit, any current that flows through one bulb must go through the other bulbs as well, so each bulb draws the same current. Since all three bulbs are 40-watt bulbs, they have the same resistance, so the voltage drop across each one is the same and equals one-third of the applied voltage, or 120/3 = 40 volts. The resistance of a light bulb filament changes with temperature, but if we ignore this, we can at least roughly estimate the current flow and power dissipation in the series circuit. We have 120 V/(360 + 360 + 360) ohms = 1/9 A. The power dissipated in each bulb is either (1/9)

^{2}× 360 = 4.44 watts, or (1/9) × 40 = 4.44 watts. The total power dissipated in the circuit is three times this, or 13.3 watts ((1/9)^{2}× 3(360) = 1080/81 = 13.3 W, or (1/9) A × 120 V = 13.3 W).With fresh light bulbs, direct measurement with an ammeter shows that the actual current flowing in the parallel circuit is 0.34 A for one bulb, 0.68 A for two bulbs and 1.02 A for three bulbs, and in the series circuit it is 0.196 A. So the current, and thus the dissipated power (23.5 watts), in the series circuit are almost twice what we arrived at above.

An “ohmic” resistance is one that stays constant regardless of the applied voltage (and thus also the current). If the light bulbs behaved this way, the measured current in the series circuit would agree with the estimate above. Even though they do not, this demonstration gives a good sense of the difference in behavior between a series and parallel circuit made with three identical resistors.

References:1) Howard V. Malmstadt, Christie G. Enke and Stanley R. Crouch.

Electronics and Instrumentation for Scientists(Menlo Park, California: The Benjamin/Cummings Publishing Company, Inc., 1981), pp. 32-32.