You have been told what equation to use so this is an "A" question.

V= IR. V is the voltage, I is the current and R is the resistance.

You have been told the current is 1.2 A and that the lamp has a resistance of 4 ohms.

Write the formula and the numbers that you will substitute in it.

V  = IR

V = 1.2 x 4     

V =  4.8 V

This time you will have to select the appropriate formula so this can be a "M" question. You are asked to calculate P of this lamp. Remember that power depends on TWO things - the current through and the voltage across the component.

You know the current through it - 1.2 A, and you have just calculated the voltage across it to be 4.8 V

P= VI
P =  4.8 x 1.2
P =  5.76 W

The units for power are watts, symbol W. Since 1W is also 1 Js^-1, you could have also answered 5.76 Js^-1

Power is the rate at which energy is converted from one form to another - this is the basic definition - LEARN IT!!

(Power can also be the rate at which energy is suppliedbut in this question we are talking about the lamp where electrical energy is being converted into heat & light energy).

A power of 5.76 W means that 5.76 J of energy are being converted into heat and/or light per second (by the lamp).

A simple answer involving a definition or simply saying 5.76 W is 5.76 Js^-1will gain you an "A" but a "M" requires you to link your answer to the energy change IN THAT LIGHT BULB.

 

This is more tricky - since you will need to work something out - to use in another calculation to find the final answer, this is a "M" and "E" question. But don't panic. Let's see what you have already got.

We know that the total voltage in the circuit is 12V - look at the diagram - the supply voltage is 12V. You have calculate that the voltage across the other light bulb (the 4 ohm one) is 4.8V. Therefore the voltage across lamp B must be 12 - 4.8 V.

12 - 4.8 = 7.2V.

So far so good. What else do you know. The current is the same "all around" a series circuit, and you have been told that it is 1.2A.

So you know the current through - and the voltage across - lamp B.

Select the formula that involves V, I and R. yes - V=IR. You will need to rearrange it to find R.

R = V/I - make sure you show that you have used this formula.

R = 7.2/1.2 - substitute the numbers you know into the formula

R = 6 ohms - don't forget the units!!

There is another way you could do this calculation and it relies on you remembering that the total resistance of 2 components in series is the sum of their resistances..... yes! That means that if you work out the total resistance of the 2 lamps, and then subtract the resistance of the first lamp (4 ohms) you will know the resistance of lamp B. Let's try!

Voltage lost across both the lamps is 12V in total. The current is 1.2A. Since R=V/I the total resistance of the 2 lamps is 12/1.2 is 10 ohms. Since the one lamp has a resistance of 4 ohms, lamp B must have a resistance of 10-4 = 6 ohms.

The more powerful the lamp, the brighter the lamp. P = V x I. Lamp B will be brigter.

Why? The question doesn't ask you to explain why - but you will have worked something out to come to your answer of "Lamp B"P=V x I. Since I is the same through both, and more voltage is lost across lamp B, then lamp B will be more powerful.

The ammeter can be placed anywhere in the circuit before the point where the current splits to go through the 2 parallel strands.

The parallel circuit has less resistance than the series circuit due to an increased number of pathways. A lower resistance means a larger current will flow.

Both bulbs receive 12V (compare this with the series circuit where the voltage was "shared"). Since V=IR and they now receive 12V but the total resistance is less ..... so this leads to this higher current.

You have calculated lamp B to have a resistance of 6 ohms. The 4 ohm lamp will be the brighter.  Since it has a lower resistance there is more current flowing through it. This results in a greater power outputthan lamp B (which has a higher resistance). They both have the same voltage across them - 12V - since they are in parallel. Since P=VI the one with the greater current through it will be more powerful - convert more energy per second - and therefore be brighter.