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

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The discussion revolves around simulating a circuit with two 12V batteries and a 6V lamp bulb to verify calculations. Key calculations include the lamp load current of 0.1667A, Thevenin resistance of 1.333 ohms, and Thevenin voltage of 12V. Participants express uncertainty about the voltage across the bulb and the implications of connecting a 6V bulb to a 12V circuit, suggesting the need for additional components to ensure proper operation. Concerns are raised about potential rounding errors in calculations, emphasizing the importance of accuracy in electrical simulations. The conversation highlights the complexities of circuit behavior when components are not used within their specified ratings.
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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
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

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,
 
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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?

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Likes Tom.G and berkeman
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.
 

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