How to Find Voltages Between Light Bulbs in a Series Circuit and Current?

As the problem is presented there are assumptions you must make to attempt this problem. The fundamental initial assumption is to assume both light bulbs act linearly (just like a resistor) at different voltages than their ratings. This is a false assumption but you must start somewhere. Perform the Kirchoff loop calculation to find the current. Now look up what the voltage drop should be for each bulb with this current. The current to voltage response will be different depending on bulb design. If this voltage drop matches your calculated number then you are done. Most likely it won't match. Substitute the voltage drop for the bulbs and recalculate the current flowing through the resistor. Apply this new current to find the next iteration of voltage drops across the bulbs. repeat. The voltage and current numbers will slowly converge to some set of values. When 100%*(In-I_n-1)/I_n-1<resistor tolerance then you are clearly done with the exercise. This technique is called successive approximation.

Remember to show all work or you should not get credit.

You are making this way too complicated.

I=V/R The current in all three items in series is the voltage drop over the 1K Ω resistor, in volts, divided by the value of the resistor 1K = 1000 Ω.

For example, if the voltage drop just happened to be 12 V, then I = 12/1000.

Thus the current would be 12 mA, in such a case. With that large a resistance, it is unlikely the voltage drop would be that low, so if the entire 24 V drops across the 1K resistor, then current would be 24 mA.

Or stick a DVM on the test points and read the voltage drops across each item, and source to ground, then it is a simple matter to understand the equivalent circuit of the lamp bulbs as connected with the source and resistor.

The voltage vs current plot for an incandescent lamp is not a straight line because when the filament gets hot, it's resistance increases dramatically.

For a resistor, the power dissipated is V^2/R. For an incandescent lamp, the power is proportional to V^1.6 = k x V^1.6, where k is a different constant for each light bulb.

https://en.wikipedia.org/wiki/Incandescent_light_bulb

Power = V x I = k x V^1.6, (divide by V)

k x V^0.6 = I is the relationship of voltage to current

For each bulb, you can find k, k = Power / Voltage ^ 1.6

bulb1 Power=10W when Voltage = 12, solve for k1

bulb2 Power=25W when Voltage = 12, solve for k2

With k1 and k2 you can plot voltage vs current for each bulb

Current through each bulb and resistor is the same. So total voltage is sum of bulb1 plot, bulb2 plot and plot of voltage across resistor versus current (V=I*R)

Sum the three plots and where the total voltage is 24 volts, that is your current.

Go back and substitute this current in the voltage versus current formula for bulb1, bulb2, and resistor.

Electronics 101:

In SERIES circuits, CURRENT is common.

In PARALLEL circuits, VOLTAGE is common.

1. Familiarize yourself with the series circuit characteristics....as a starting point!

2. The function generator may not have enough power to drive your series loads!

3. Tools if available may not be useful to measure any voltage drops since the series loads since the bulbs have a much higher power ratings?

4. Tendency of Higher power rated loads is to draw higher current which in turn will pull the output of a multifunction generator down to zero!

if this the case, all you can do is just make assumptions?

On the assumption that this is not homework, you measure each voltage with a voltmeter. Duh

An important note is: Light bulb's "cold" resistance will be different than "Hot" resistance. Cold resistance will be several times lesser than hot resistance.

Source: Wiki

Current and resistance

The actual resistance of the filament is temperature dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms).^[57][107]Since incandescent lamps are resistive loads, simple phase-control TRIAC dimmers can be used to control brightness. Electrical contacts may carry a "T" rating symbol indicating that they are designed to control circuits with the high inrush current characteristic of tungsten lamps. For a 100-watt, 120-volt general-service lamp, the current stabilizes in about 0.10 seconds, and the lamp reaches 90% of its full brightness after about 0.13 seconds.^[108]

Carbon filament bulbs have the opposite characteristic. The resistance of a carbon filament is higher when it is cold than when it is operating. In the case of a 240 Volt, 60 Watt carbon filament bulb, the resistance of the filament when at operating temperature is 960 Ohms, but rises to around 1500 Ohms when cold (

From information provided by Lyn, these are the meters provided to give you the ability to measure the voltages/currents in the network you set up, using the values and components within the Multisim library, an exercise in Ohm's law.

The rest of the fine answers given here are based on the reality of the engineering task presented to you, not as simple as was likely intended, an unrealistic assembly of components with highly questionable functional usefulness. However, the intent was to get you to use the meters/instruments after building a simulation. Who knows just what kind of lamp is in your library, for instance?

4. Part C: Simulating the Circuit

You are now ready to run an interactive Multisim simulation; however, you need a way to visualize the data. Multisim provides instruments to visualize the simulated measurements. Instruments can be found on the right menu bar and are indicated by the following icons.

Figure 7. Instruments toolbar