# Physics power question frustrating me

1. Mar 27, 2010

### thereddevils

1. The problem statement, all variables and given/known data

Two bulbs P and Q labelled 12V , 24 W and 12 V and 36 W respectively are connected in series with a battery 24 V . Among the following statements , which one describes correctly about bulb P and Q

(a) potential difference across P > pd across Q

(b) resistance of Q > resistacne of P

(c) current passing through Q > current in P

(d) power dissipated from Q > power dissipated from P

(e) potential difference and current in both bulbs are the same

3. The attempt at a solution

For (a) , i don think its right , since from P=VI , where I is constant here , when P increases ,V also increases .

For (b) , i think its correct because from P=I^2 R , when P increases , R also increases .

(c) its not correct since the current passing through both bulbs are the same

(d) this is also correct , since given that Q is 36 W and P is 24 W

(e) this is incorrect .

Is my reasoning for each correct ?

2. Mar 27, 2010

### kuruman

Re: power

What does one mean when one says "a light bulb is labeled 12 V and 24 W"? What information can be extracted from this?

3. Mar 27, 2010

### thereddevils

Re: power

the potential difference across it is simply 12 V and the power it requires to light is 24 W , is it that simple ? And also the current is 2 A .

4. Mar 27, 2010

### kuruman

Re: power

Not quite. A light bulb will give some light even if the applied voltage is less than the rated voltage. It means that when 12 V is applied to the bulb, the power dissipated in it will be 24 W. Can you use this information to find the resistance of the bulb?

5. Mar 27, 2010

### thereddevils

Re: power

oh , yeah , P=V^2/R so when the power dissipated from the bulb is higher when the same amount of voltage is supplied to the bulb , the resistance will be higher .

so (b),(e), (c) are out .

is (d) correct ?

6. Mar 27, 2010

### kuruman

Re: power

Why don't you calculate the power dissipated in each resistor and answer your own question?

7. Mar 27, 2010

### thereddevils

Re: power

ok , thank you for your help .