Create a circuit in simulation to prove my work (2 batteries and 1 lamp bulb)

so this is a question i am trying to solve and then confirm with the simulation,
so far i have..
Lamp load
IL = PL/VL 1/6 = 0.1667A

Thevenin resistance Rth = Rs1 || Rs 2 = 2 x 4 / 2 +4 = 1.333ohms

Vth = 12V

look at node voltage with load circuit and use kcl
V = 6V

Is1 = 12 -6 / 2 = 3A

Is2 = 12-6/4 = 1.5A

supply can be 4.5A but lamp load is 0.1667A

i think the resistance of the light source is 35.99 ohms ?

i am not sure how i draw this to confirm my working out, i thought i would recreate the circuit and the amp meter would read 4.5A going into the lamp,

It seems odd that they would specify a 6V lamp bulb to use in a 12V circuit. Are you sure you copied the problem statement accurately?

Perhaps by saying "create a circuit" they mean to add other components to make the circuit function within specifications? A resistor perhaps?

i have the details correct

leejohnson222 said:

Homework Statement: create a circuit with 2 dc sources of 12v in parallel, one battery has a resistance of 2ohms the second has 4ohms, a lamp is connected between terminal A and B with a 6V rating and 1W
Relevant Equations: Thevenin resistance Rth = 1.333ohms Not given equations - the results of your calculation?
Thevenin Voltage Vth = 12V

so this is a question i am trying to solve and then confirm with the simulation,
so far i have..
Lamp load
IL = PL/VL 1/6 = 0.1667A IF it is supplied with 6 V.

Thevenin resistance Rth = Rs1 || Rs 2 = 2 x 4 / 2 +4 = 1.333ohms

Vth = 12V

look at node voltage with load circuit and use kcl
V = 6V But you don't yet know what the voltage across the bulb is. (see below).

Is1 = 12 -6 / 2 = 3A Likewise, you don't know the terminal voltage of the battery.

Is2 = 12-6/4 = 1.5A

supply can be 4.5A but lamp load is 0.1667A

i think the resistance of the light source is 35.99 ohms ? Beware of rounding errors. (see below)

i am not sure how i draw this to confirm my working out, i thought i would recreate the circuit and the amp meter would read 4.5A going into the lamp,

Sorry, I can't help with your question, but IMO that is impossible without more information. But You may be interested to know:
You cannot assume that a "6 V, 1 W bulb" will have 6V pd across it, nor that it will pass 1/6 A when connected to anything other than a supply which puts 6V across it.

A lamp rated as 6 V 1 W means, if connected to a 6V supply it should use 1 W and hence draw 1/6 A, so its resistance would thern be 36 Ohm. But if you connect it to a supply producing a different voltage, then there may be a different voltage across the bulb.
So a supply putting 3 V across the bulb, might pass only 1/12 A giving power of only 3/12 or 1/4 W.

(In fact, a light bulb, because the filament heats up so much, usually has a much lower resistance when it is cold. You can't easily calculate that resistance, but you know it will be less when the light is off or dim, so at 3 V it might have a resistance of (eg.) 30 Ohm, so current 3/30 = 1/10 A and power = 3 x 0.1 = 0.3 W )

PS.
Beware of rounding errors. (unimportant here, but can be.)
IF you calculate 6.000 / 0.1667 = 35.99280 etc. then, because you used a value rounded to 4 sf, you can rely on only 3sf, which = 36.0 which is the answer you get if you don't round 1/6 to 0.1667. Work to one more place than you give your result. 35.99 is fine as an intermediate answer - still working with 4sf, but as a final result, round it to 36.0 to 3sf.