Physics doubts. Urgent help is needed!?

Question

1) There are two equations for power consumed by a device:

P = I²R
P = V²/R

The first expression indicates that Power is directly proportional to R whereas the second expression indicates inverse proportionality.
How can the seemingly different dependence of P on R in these expressions be explained?



2) An electrician puts a fuse of rating 5A in that part of domestic electrical circuit in which an electrical heater of rating 1.5kW, 220V is operating. What is likely to happen in this case and why? What change, if any, needs to be made?

Answer 1

From ohm's law,
I = V / R, V = I * R, R = V / I
The equation for power, P = Voltage x Current = V * I
But V = I / R, & substitute the above value in above eqn
P = V * I = (I * R) * I = I² * R ...............(1)
P = V * I = V * (V /R) = V² / R............(2)
In eqn (1), P is directly proportional to I² & R
In eqn (2), P is directly proportional to V² & inversely to R.
Comparing to the values of V & I , V is much much higher than that of I. Hence, the difference in their squares will be still higher. So in the second equation ,the extra magnitude caused by squaring the V will be neutralised by inverse proportionality of the resistance.

(2) The rating of the heater = 1.5 kw = 1500 w & 220 V
As Power P = V * I, I = P / V
Hence, Current = 1500 / 220 = 6.8 Ampere.
As the fuse is only of 5 A the fuse will blow off. To protect this any value higher than 6.8 Amps ( as a safety margin make it 7 amps)

Answer 2

1) You can't look at the R separate from the rest of the expression. Within V, I, and R, exactly two of them are dependent on each other, as the olm law sets the constraint V=IR.

You need to assume either I or V is constant (independent) with R changing. If I is constant, then it's proportionality. If V is constant, it's inverse proportionality. Depends on if you're using a voltage source (constant V) or current source (constant I). So it's indeed not a conflict.

2) P=VI. or 1500=220*I, leading to I>5A. So the current is expected to be stronger than the fuse is allowed to carry. Therefore, replace the fuse with one with a higher rating, say, 10A.

Answer 3

Power is actually V I

In the first equation group one of the I's and the R to get V so that P = V I in the second equation group one of the V's divided by R to get I so again we get P = V I

Or in other words for the first equation as R increases the V must increase and therefore power, the second equation as R increases the I will decrease thereby decreasing power.

2) 1.5kW / 220 V would require at least a 7 Amp Fuse. I would probably round it up to 10 Amps.

Answer 4

Here is my answer. Not sure if it is 100% though.

Voltage = amperes x resistance (V = IR) so V^2 = I^2 x R^2 and I^2 x R^2/ R still leaves R in the direct proportionality relation, (I^2 x R) = P.

Answer 5

Voltage = amperes x resistance (V = IR) so V^2 = I^2 x R^2 and I^2 x R^2/ R still leaves R in the direct proportionality relation, (I^2 x R) = P.

Answer 6

1) You can think about it this way: If the current stays the same and you increase the resistance, the voltage (and thus the power) must increase in order to force the same current through a higher resistance. If the voltage stays the same and you increase the resistance, then less current flows so the power decreases (and vice versa).

2) P = IV is what you apply to solve this problem.

Answer 7

In the formulae, I and V are not constant terms. In fact I and V depend on R. So you can't say P is directly propotional to R.

I =Sqrt(P/V)

= sqrt(1500/220)

Answer 8

I can answer the first question.
Directly proportional means P = (constant) R. I is not a constant so you cannot say power is directly proportional to R.

Answer 9

1. The resistor and filament are in parallel, so there is always a circuit if the filament burns out. 2. The combined resistance of filament and parallel resistor is lower than that of the resistor alone. This means that the potential difference across the resistor will rise when the filament burns out. The consequence is a reduction in PD across the other bulbs in circuit. Therefore the chain of lights will dim, when one filament burns out.