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konstantin123 [22]
3 years ago
13

When a resistor is connected to a 12-V source, it draws a 185-mA current. The same resistor connected to a 90-V source draws a 1

.25-A current. (a) Is the resistor ohmic? Justify your answer mathematically. (b) What is the rate of Joule heating in this resistor in both cases?
Physics
2 answers:
Svetach [21]3 years ago
8 0

Answer:

The resistor is not Ohmic

Explanation:

In the first case, from ohm's law: V=IR

R=V/I= 12/0.185=64.9 ohm

In the second case:

R=V/I= 90/1.25= 72ohm

The relationship between voltage and current is linear according to ohms law but that is not the case here.

Joule heating = I^2R

For case1: (0.185)^2 × 64.9 = 2.2W

For case2: (1.25)^2 ×72 =112.5W

topjm [15]3 years ago
7 0

Answer:

Explanation:

Ohm's law is define by the formula

V = IR where I is current in A, R is resistance in ohms

to calculate the resistance we use the formula above

I = 185 mA / 1000 = 0.185 A

case 1, R = V / I = 12 V / 0.185 A = 64.86 ohms

case 2, R = V / I = 90 / 1.25 A = 72 ohms

a) the resistor is not ohmic since there is no linear relationship between R and V ( as voltage increases, resistance ought to also increase)

b) rate of Joule heating in the case 1, P, power = IV = 0.185 A × 12 = 2.22 W

case 2, P = IV = 1.25 A × 90 V = 112.5 W

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a) F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

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a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

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F cos \theta - \mu_k R = 0 (2)

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And substituting into (2)

F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

b) \mu_s=cot(\theta)

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

R-Fsin \theta - mg = 0 (1)

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Therefore, using the same procedure as before,

R=Fsin \theta +mg

F cos \theta - \mu_s (Fsin \theta +mg) = 0

And solving for \mu_s,

F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}

Now we analyze the expression that we found. We notice that if the force applied F is very large, F sin \theta >> mg, therefore we can rewrite the expression as

\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)

So, this is the critical value of the coefficient of static friction.

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