two 28volt light bulbs

...there is no such thing as a 28volt light bulb. You might say it is a bulb suitable for a circuit with a 28volt emf. The bulb will have a specification that includes maximum power in watts and luminescence.

 

...there is a confusion caused by language as the emf in a circuit is called "voltage" the same word as the "voltage" drop across a resistor but they are not the same thing.

But they measure the same.

 

But they measure the same.

here's an exam question for you...
Explain the difference between the following terms:
a) voltage emf
b) voltage drop
c) voltage measurement

 

What controls Voltage Drop?
1- If you have a single bulb 28volt light circuit the voltage drop thru the bulb is 28 v (in essence all the voltage is consumed thru the bulb).

2- If you have a series circuit with two 28volt light bulbs, the voltage drop at the first bulb is less than 28 v and the remaining voltage passes on to the second bulb in the series (how does the voltage know to go past the first bulb and what tells some of the voltage to go on to the second bulb in series). This all happens very fast, there must be some triggering mechanism that alerts the Voltage that there is a second bulb.

How does water know to flow down hill?
Current isn't conscious and doesn't "know" anything.

Resistance determines voltage drop.

 

here's an exam question for you...
Explain the difference between the following terms:
a) voltage emf
b) voltage drop
c) voltage measurement

Sorry, I gave up doing exams about 50 years ago.
But they do all measure the same way with the same parameter which is all that is of interest to me.
The semantic differences are not confusing to me, although I suppose it might be to a neophyte.
And I see we now seem to have another pedantic poster on these forums.

 

An interesting question.

Us old ones can do with the odd kick like this,

Can I suggest we talk about resistors not light bulbs, as bulbs are another level of complexity that is not relevant.

Think one resistor , a switch and a voltage source ( say a battery ) in series.
Switch open, zero volts across the resistor.

Now close switch.

How does the battery know how much current to send ?

it dosn't.

You close the switch, and the current flows, the voltage across the resistor (quickly) rises. as the current rises.
as the voltage across the resistor rises, the resistor, by ohms law, drops a voltage,

the final current is defined by the resistance of the circuit.

Eventually, within pico seconds, the full battery voltage is applied across the resistor, and things are stable.

if you have a fast enough scope, you can see this happening...

now cut that resistor in two, wire in series and try again.

same total resistance, so with same same total final current will flow.

We know that the same current must flow through both resistors, as they are in series and current can not go around,

same total resistance, same current, same total voltage drop.

the actual voltage across each resistor will be proportional to the resistance / current flowing, but as the same current flows in both resistors, in normal to think of the voltage being divided in the resistors ratio, a voltage divider.

The key is to follow the current.

 

Hi there,

In short, when the voltage is first applied a field is set up that starts the electrons into motion.
It is believed that there is no delay between the time the voltage is first applied and the time the
electrons start to move, so it is considered instantaneous, although they may move only a little at
first as the voltage starts to build up. The field travels at the speed of light, but the electrons
move rather slowly, and they move slower through higher resistance materials. The two resistances
act as bottlenecks however keeping the pace of each electron at some slower rate, slower than if the
resistance was low like in a copper wire.

Think about an ideal case, like a wire that is 1 atom in diameter compared to a wire that is 2 atoms in diameter, both the same length, in series, see what you can come up with.

Alternately think about another ideal case, where we have 1 wire in series with 2 more wires of the same length as the first, but the 2 more wires are in parallel, and the entire combo is in series. So that is 1 wire in series with 2 wires where the 2 wire pair is in parallel, and so forms a single wire with 2 times the area of the 1 wire wire.
This would look like:
---------------==========

and both the 1 wire wire is unit length and the 2 wire wire is unit length.
The 1 wire wire will have 1/2 the resistance of the 2 wire wire.
See what you can come up with using either one of these ideal cases

In simulation, we could look at two resistors in series with an inductor driven by a battery and observe the voltages across the two resistors. The inductor acts much like the wire because it is like a mostly inductive transmission line. In fact, to find out how it works in more detail you might look into transmission lines and how the energy distributes over them with time and distance from the source.

 

The sum of the I*R drops is equal to the applied voltage.
This means the components AND the drops across the wire segments.

This is a variation of Kirchoff's Voltage Law where the sum of the IR drops in a loop equals zero.

The light bulb us a current dependent resistor with manufacturing tolerances.