How do i work out the voltage an current through these lamps?

In a series circuit you add up the resistances. So, one 12ohm resistor and 2 12ohm lamps gives you 36 ohms in total. As the resistances are equal the voltage across each will be equal, and as there are 3 of them at 15v there'll be 15/3 which is 5V.

Current anywhere is the circuit is equal, and as your total resistance is 36 ohms you can use Ohm's law to find the current in the circuit.

Ohm said V=IR and therefore I(current) = V/R. Substituting numerical values I= 15/36 and so 0.42 A or 420 milliamps, and that's the current that's going to be in each bulb.

In a parallel circuit the resistances are calculated thus 1/Rt= 1/R1+1/R2 +1/R3 and so on.

Your 12 ohm bulbs are in parallel and so their total resistance = 1/12+1/12 or 2(1/12). If you get your calculator out it comes to 0.16666

and 1/0.16666= 6. Your bulb resistance is therefore 6 ohms and you can count the total resistance of the parallel as one resistor. In effect what you have now is two resistors in series, one 12 ohm and one 6. So add them up: 18 ohms.

Using Ohm again: I= V/R I= 15/18 = 0.8333A

Now, in a series circuit you divide the voltage but not the current.

In parallel, you divide the current but not the voltage.

This bit's hard so read slowly..................Kirchoff's Law says that the current at a junction is equal to the currents flowing in the entrance or exit to that junction. You have two bulbs, and the current is 0.83 amps, and so the current flowing through each branch of that junction must be 0.83/2, which is 0.415 , and that's near enough to the 0.42 Amps in the series circuit to prove the same current's flowing.

Now we need to talk about power, which is measured in Watts.

Power (W) = Volts * Amps.

We know that the current is constant in both circuits (0.42A, remember) BUT:

In the series circuit the bulbs have 5 volts each, and 5*0.42= 2.1 Watts

In the parallel, the bulbs have 15 volts each and 15*0.42 = 6.3 W.

The more power developed over the bulb the brighter it shines (unless you put more power in than it can handle in which case it blows) and as you can see, the bulbs in the parallel will shine threefold more brightly than the series.

In this answer you've done half a term of basic electrical theory, so don't worry if you don't get it first go. Plug away reading it again until you do.

A) Voltage (V) is present day (I) situations resistance (R), via Ohm's regulation. So a hundred and twenty=.4(R) a hundred and twenty/.4=R R=3 hundred Ohms B) chilly products often have decrease resistance because of the fact the molecules interior are shifting greater slowly. consequently: a million/5 situations 3 hundred=60 Ohms. C) returned, notice Ohm's regulation (V=IR). V=a hundred and twenty I=unknown R=60 so, a hundred and twenty=60(I) a hundred and twenty/60=I=2 Amps.

Two elements are connected in series if the current passing through one element must pass through the other. They are in parallel if there are two paths possible.

For three items (resistor and two light bulbs) there are three combinations possible:

1) All three in series

2) Two in parallel connected to the third in series

3) All three in parallel

You said that the two lamps are in parallel with each other . Is the resistor in series with them or in parallel to them?

Follow the rules of Ohms Law and the basic mathematical practices

There is a calculator on this web page to help

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Que !!?