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You are watching: A 100-w light bulb is operated by a 110-v dc source. what is the resistance of this bulb?

OK. These room actually AC circuits. Due to the fact that the lots are practically purely resistive, i.e., there are no capacitances or inductances (or they are tiny enough to be negligible), and also since the rms (root-mean-square) AC voltage and also current act in purely resistive circuits as DC voltage and current do, the two circuits shown over are tantamount to the matching DC circuits. The AC indigenous the wall is sinusoidal. The rms voltage because that a sinusoid is 0.707Vp, whereby Vp is the peak voltage. Similarly, the rms current through a resistor is 0.707ip, wherein ip is the top current. These effective worths correspond come the DC values that would offer the very same power dissipation in the resistor. These space slightly various from the average voltage and current, which space 0.639Vp and also 0.639ip for a sinusoid. For AC from the wall, the rms voltage is roughly 120 V, and the typical voltage is about 110 V.

Each board has actually three 40-watt bulbs, associated as presented by the resistor circuit painted ~ above 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 pear draws 1/3 A. (The devices in iV space (C/s)(N-m/C), or J/s, which are watts.) because that a offered resistance, V = iR, for this reason the bulb’s resistance (when it has actually 120 volts throughout it) is 120/(1/3), or 360 ohms. (We additionally know by the 2 equations above that ns = i2R, which offers R together 40/(1/9), or 360 ohms.)

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

In the collection circuit, any current that flows through one bulb have to go through the various other bulbs as well, so each pear draws the same current. Since all 3 bulbs room 40-watt bulbs, they have the same resistance, therefore the voltage drop across each one is the same and also equals one-third of the applied voltage, or 120/3 = 40 volts. The resistance that a light bulb filament transforms with temperature, but if we neglect this, we have the right to at least about 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 bulb is one of two people (1/9)2 × 360 = 4.44 watts, or (1/9) × 40 = 4.44 watts. The complete 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 v an ammeter shows that the actual current flowing in the parallel circuit is 0.34 A because that one bulb, 0.68 A for 2 bulbs and 1.02 A for 3 bulbs, and also in the series circuit it is 0.196 A. For this reason the current, and also thus the dissipated strength (23.5 watts), in the series circuit are nearly twice what we arrived on above.

An “ohmic” resistance is one the stays constant regardless of the applied voltage (and thus additionally the current). If the irradiate bulbs behaved this way, the measured current in the series circuit would agree v the estimate above. Even though they do not, this demonstration gives a an excellent sense of the distinction in actions between a series and parallel circuit made v three the same resistors.

What wake up if the irradiate bulbs are not all of the very same wattage rating?

An interesting variation of this demonstrate is to display what happens once we put light bulbs that three various wattages in each circuit. A an excellent choice is to keep one 40-W light bulb in every circuit, and then include a 60-W bulb and also a 100-W bulb. In the parallel circuit, as noted above, the voltage throughout each pear is the very same (120 V), so each bulb draws the existing that it would certainly if that alone were connected to the wall, and the intensities the the bulbs for this reason vary as you would mean from the wattage ratings. The 100-W bulb is the brightest, the 40-W bulb is the dimmest, and also the 60-W bulb is what in between. When we put the same mix of bulbs in series, an interesting thing happens. Since both the 60-W bulb and the 100-W bulb have lower resistance 보다 the 40-W bulb, the existing through the circuit is somewhat greater than for the 3 40-W light bulbs in series, and the 40-W bulb glows much more brightly 보다 it did as soon as it to be in series with two various other 40-W bulbs. The current through this circuit procedures 0.25 A. This is around 76% the the 0.33 A that the 40-W bulb would attract by itself, fifty percent the 0.5 A that the 60-W bulb would draw, and also 30% that the 0.83 A the the 100-W bulb would certainly draw. In ~ this current, the 40-W bulb lights relatively brightly, the 60-W bulb simply barely glows, and also the 100-W bulb does not light in ~ all. The photograph listed below shows the operation of these 2 circuits:


The bulbs in each circuit, from left come right, room a 40-W, 60-W and a 100-W light bulb. In the parallel circuit, the bulbs obviously boost in brightness indigenous left come right. In the collection circuit, the brightness decreases from left come right. The measure up voltages in the circuit room 120 V throughout all three bulbs, 109 V throughout the 40- and also the 60-W bulbs, and 78 V across the 40-Watt bulb. The voltage drop throughout the 60-W bulb is thus 31 V, and also it is 11 V throughout the 100-W bulb. Multiplying every of these by the 0.25-A current, we uncover that in the series circuit, the 40-W pear dissipates around 20 watts, the 60-W bulb dissipates 7.8 watts, and also the 100-W bulb dissipates about 2.8 watts, which synchronizes with the family member intensities us observe for the 3 bulbs.

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