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OK. These space actually AC circuits. Since the loads are nearly purely resistive, i.e., there room no capacitances or inductances (or lock are tiny enough to be negligible), and also since the rms (root-mean-square) AC voltage and also current act in completely resistive circuits as DC voltage and current do, the 2 circuits shown over are equivalent to the equivalent DC circuits. The AC from the wall is sinusoidal. The rms voltage for a sinusoid is 0.707Vp, whereby Vp is the top voltage. Similarly, the rms current through a resistor is 0.707ip, where ip is the optimal current. This effective values correspond come the DC worths that would provide the exact same power dissipation in the resistor. These are slightly different from the average voltage and current, which room 0.639Vp and 0.639ip because that a sinusoid. Because that AC indigenous the wall, the rms voltage is roughly 120 V, and the average voltage is around 110 V.

Each board has three 40-watt bulbs, connected as displayed by the resistor circuit painted top top it. The board on the left has actually the bulbs arranged, that course, in parallel, and the board on the right has them in series. Since power, P, equates to iV, P/V = i, so in ~ 120 V, a 40-watt bulb draws 1/3 A. (The systems in iV space (C/s)(N-m/C), or J/s, which space watts.) because that a given resistance, V = iR, for this reason the bulb’s resistance (when it has actually 120 volts across it) is 120/(1/3), or 360 ohms. (We additionally know by the 2 equations over that ns = i2R, which provides R together 40/(1/9), or 360 ohms.)

When the bulbs are linked in parallel, each bulb has actually 120 V throughout it, each draws 1/3 A, and each dissipates 40 watts. In this circuit, all bulbs glow in ~ their complete brightness. The complete power dissipated in the circuit is 3 times 40, or 120 watts (or 3(1/3) A × 120 V = 120 W).

In the series circuit, any current that flows with one bulb need to go with the various other bulbs together well, so each pear draws the same current. Since all three bulbs are 40-watt bulbs, they have actually the exact same resistance, so the voltage drop across each one is the same and also equals one-third of the used voltage, or 120/3 = 40 volts. The resistance of a light pear filament alters with temperature, however if we neglect this, we can at least around estimate the current flow and power dissipation in the series circuit. We have 120 V/(360 + 360 + 360) ohms = 1/9 A. The strength dissipated in each pear is either (1/9)2 × 360 = 4.44 watts, or (1/9) × 40 = 4.44 watts. The full power dissipated in the circuit is 3 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 irradiate bulbs, direct measurement through an ammeter reflects that the actual existing flowing in the parallel circuit is 0.34 A for one bulb, 0.68 A for 2 bulbs and 1.02 A for three bulbs, and also in the series circuit the is 0.196 A. So the current, and also thus the dissipated power (23.5 watts), in the series circuit are nearly twice what we landed on above.

An “ohmic” resistance is one that stays consistent regardless of the applied voltage (and thus likewise the current). If the irradiate bulbs behaved this way, the measured current in the collection circuit would certainly agree v the calculation above. Even though they do not, this demonstration offers a an excellent sense that the distinction in habits between a series and parallel circuit made with three the same resistors.

What happens if the irradiate bulbs room not all of the same wattage rating?

An exciting variation of this demonstration is to display what happens when we placed light bulbs the three various wattages in every circuit. A great choice is to keep one 40-W light bulb in each circuit, and then add a 60-W bulb and also a 100-W bulb. In the parallel circuit, as detailed above, the voltage throughout each pear is the exact same (120 V), therefore each bulb draws the present that it would if it alone were connected to the wall, and the intensities the the bulbs therefore vary together you would mean from the wattage ratings. The 100-W pear is the brightest, the 40-W pear is the dimmest, and the 60-W bulb is somewhere in between. As soon as we placed the same mix of bulbs in series, an exciting thing happens. Because both the 60-W bulb and also the 100-W bulb have lower resistance 보다 the 40-W bulb, the present through the circuit is somewhat greater than because that the three 40-W light bulbs in series, and the 40-W bulb glows more brightly than it did as soon as it to be in series with two various other 40-W bulbs. The present through this circuit measures 0.25 A. This is around 76% the the 0.33 A that the 40-W pear would draw by itself, fifty percent the 0.5 A that the 60-W bulb would draw, and 30% the the 0.83 A that the 100-W bulb would draw. In ~ this current, the 40-W pear lights fairly brightly, the 60-W bulb simply barely glows, and also the 100-W bulb does no light in ~ all. The photograph below shows the procedure of these 2 circuits:


The bulbs in every circuit, indigenous left come right, room a 40-W, 60-W and a 100-W irradiate bulb. In the parallel circuit, the bulbs obviously rise in brightness native left to right. In the series circuit, the brightness decreases native left to right. The measured voltages in the circuit are 120 V across all three bulbs, 109 V across the 40- and also the 60-W bulbs, and 78 V throughout the 40-Watt bulb. The voltage drop across the 60-W pear is hence 31 V, and it is 11 V throughout the 100-W bulb. Multiplying each of this by the 0.25-A current, we uncover that in the series circuit, the 40-W pear dissipates about 20 watts, the 60-W bulb dissipates 7.8 watts, and also the 100-W bulb dissipates about 2.8 watts, which coincides with the relative intensities we observe because that the three bulbs.

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1) Howard V. Malmstadt, Christie G. Enke and Stanley R. Crouch. Electronics and Instrumentation for Scientists (Menlo Park, California: The Benjamin/Cummings posting Company, Inc., 1981), pp. 31-32.