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)

 

[sardar]

The power of a resistor in a parallel circuit is determined by the voltage and current passing through it. In a parallel circuit, the voltage is the same across all components, but the current is divided between the components. Therefore, when calculating the power of a specific resistor in a parallel circuit, we must take into account the portion of the total current that is passing through that resistor.

In this case, using R/2 for the power of resistor R takes into account the fact that only half of the total current is passing through that resistor. This is because the current is divided evenly between the two parallel resistors, so each resistor only receives half of the total current. Thus, using R/2 allows us to accurately calculate the power of resistor R in the parallel circuit.

If we were to use R instead, we would be assuming that the full current is passing through resistor R, which is not the case in a parallel circuit. This would result in an inaccurate calculation of the power of resistor R.