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

• Engineering
• leejohnson222
leejohnson222
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..
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?

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..
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.

berkeman

## How do I set up a basic circuit with 2 batteries and 1 lamp bulb in a simulation?

To set up a basic circuit with 2 batteries and 1 lamp bulb in a simulation, you need to connect the positive terminal of the first battery to the positive terminal of the second battery. Then, connect the negative terminal of the second battery to one terminal of the lamp bulb. Finally, connect the remaining terminal of the lamp bulb back to the negative terminal of the first battery, completing the circuit.

## What type of batteries should I use for the simulation?

The type of batteries you use for the simulation depends on the specifications of the lamp bulb. Common choices are AA, AAA, or 9V batteries. Ensure that the combined voltage of the batteries matches the voltage requirements of the lamp bulb to avoid damage or insufficient power.

## What software can I use to simulate this circuit?

There are several software options available for simulating circuits, such as Tinkercad, LTspice, Multisim, and CircuitLab. These tools allow you to create and test your circuit virtually before building it physically.

## How do I know if my circuit is correctly set up in the simulation?

In the simulation software, once you have connected all components, you can run the simulation to see if the lamp bulb lights up. If it does, your circuit is correctly set up. If it doesn't, check all connections for any mistakes, ensure the batteries are correctly oriented, and verify that the total voltage matches the lamp bulb's requirements.

## What should I do if the lamp bulb does not light up in the simulation?

If the lamp bulb does not light up in the simulation, double-check all connections to ensure they are correct. Verify that the batteries are properly oriented and that their combined voltage matches the lamp bulb's voltage requirements. Additionally, check for any software-specific settings that might affect the simulation's outcome.

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