Current of electricity- power of a resistor in parallel

[vadevalor]

I got the answer by comparing I2R of the circuit with resistor R but why do i have to use R/2 for power of resistor R?( P= (0.5I)^2 (R/2)) I know its parallel to Resistor Q but i want to find the power of that component so shouldn't it be P= (0.5I)^2 (R) instead?

 

[Simon Bridge]

shouldnt it be P= (0.5I)^2 (R)

if R is the resistance of resistors P Q and R (unfortunate notation!) and I is the total current drawn from the battery ... that's what I'd have thought. What made you think otherwise?

 

[CWatters]

Yup. Resistors R and Q have half the current of P so the powers are

P: I^2 R
R & Q: (I/2)^2 R = 1/4 I^2 R

So the ratios are 2:2:8 (or if you prefer 2+2+8=12W)

Multiple choice answer A) 2W.

why do i have to use R/2 for power of resistor R

Who says you do?

 

[Simon Bridge]

It could be a misread or a typo in a model answer - if the argument was that the power dissipated in the parallel resistors is (I^2)(R/2) then the power dissipated in one of them must be 0.5(I^2)(R/2). Somehow the 0.5 got included in the brackets?

(0.5I^2)(R/2)=(I^2)R/8 ... which would give a power not on the multiple choice list wouldn't it?

But I don't see OPs answer or all the working on the attachment ... I see the start of a sum that goes off the edge of the pic, and an arrow to the simplified result:
12W=(I^2/2)(2R/3)