Resistance questions?

Question

Hey if u could explain these to me:

1) "Lisa finds that when she increases the voltage across an ohmic resistor from 6 V to 10 V the current increases by 2 A.
What is the resistance of this resistor?"
OK so if u work out the gradient that's 4/2 = 2. The answer gives 2 ohms, but isn't resistence the inverse of the gradient, so it should be 1/2 ??

2) The resistance of a certain piece of wire is found to be 0.8 ohm. What would be the resistance of a piece of wire of twice the diameter?
if it's twice then shouldn't u do 0.8/2 = 0.4? But my book says it's 0.2 ohm.

Answer 1

1.
When Lisa did the experiment, she would plot voltage, the independent variable (The thing you change), on the x-axis, and current, the dependent variable (The thing you measure), on the y-axis.

This gives the gradient as 2A/4v = 0.5 S As you point out, the resistance is the inverse of this gradient =2 ohm

2. Resistance is inversely proportional to (cross sectional area) Area is proportional to d^2, not d.

Answer 2

Gradient is the ratio of change in y and change in x in a graph

It is not the definition of resistance.

If a graph is drawn with voltage across x axis and current in the y axis, then the slope of the straight line or gradient is ∆ y / ∆ x where∆ y & ∆ x are the change in current and change in potential.

Therefore the gradient or slope is the reciprocal of the resistance.

If a graph is drawn with current across x axis and voltage in the y axis, then the slope of the straight line or gradient is ∆ y / ∆ x where∆ y & ∆ x are the change in voltage and change in current.

Therefore the gradient or slope is the resistance.

Since in the given problem the change in potential is 4v and change in current is 2A, the resistance is 4/2 = 2 ohm.

Resistance = voltage / current.

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In the second problem, it must be stated that the length is kept constant.

We suppose that the length is kept constant.

In that case the resistance is inversely proportional to the area of cross section or square of the diameter.

R1 = K / d 1^2 where K is some constant.

R2 = K / d2 ^2

R1 / R2 = [d2 /d1] ^2

It is given = [d2 /d1] = 2.

Therefore

R1 / R2 = 2^2 = 4.

R1 = 4 R2

Or

R2 = 1/4 of R1 = 0.8 / 4 = 0.2 ohm.

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Answer 3

1) Your first answer is correct. At 6v the current flowing in a 2 ohm resistor would be 3A and at 10v the current in the same resistor would be 5A, which is a rise of 2A as the question says. If it had been 1/2 ohm then the current at 6v would be 12A and at 10v 20A - not a difference of 2A! Therefore, your second premise must be incorrect.

2) The resistance of a wire is proportional to its cross sectional area. This is given, in a circular wire by pi*r^2. Thus, if the diameter is doubled, the area goes up by 4 times, this accounts for the answer you have of 0.2.

I hope that this helps.

Dave

Answer 4

1. And another way to solve it:

R = V / I
R = 6 / I
R = 10 / (I +2)

Now set each R value equal to each other:

6 / I = 10 /(I+2)

Cross multiply an solve for I:

6I + 12 = 10 I
4I = 12
I = 3 amps

Now subsitute I=3amps into the orginal equation and solve for R.

R= V/I
R= 6V/3A
R = 2 ohms

2. R is proportional to length and inversely proportional to cross sectional area. Since you doubled the diameter, you doubled the radius of the wire, and area = pi X radius^2, therefore the area is 4 times larger, and the resistance is 1/4.

Answer 5

1) E=IR, R=E/I=dE/dI. So when the gradient is expressed this way, R is proportional to the gradient. You may have been thinking of the gradient dI/dE, which is equal to the conductance or inverse resistance.
2) R is proportional to length and inversely proportional to cross-sectional area. So doubling the diameter quadruples the surface area and resistance decreases to 1/4 its original value.

Answer 6

1) Let the resistance be R.
So 10/R - 6/R = 2
R = 2 ohms

2) Diameter is doubles means radius is doubled.
Now R = rho* L/A = rho * L / pi.r^2
So R is proportional to 1/r^2 since the rest is constant.

Now r' = 2r
So R' is proportional to 1/r'^2 = 1/4r^2
R / R' = 4

R' = R / 4 = 0.8/4 = 0.2 ohms

Hope this helps.

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