Question

A wire is made from a material having a temperature coefficient of resistivity of 0.0003125 (°C^-1). In an experiment, we maintain a constant potential difference of 150 volts across the wire and measure the power dissipation in it as a function of temperature ; at a temperature of 20° C, the power dissipation is 300 watts. What is the (%) change in the power dissipation as the wire's temperature rises from 20°C to 1820°C?

Answer 1

Answer:

The percentage change in the power dissipation as the wire's temperature rises from 20°C to 1820°C is 36%.

Explanation:

Given that,

Temperature coefficient of resistivity = 0.0003125

Potential difference = 150 V

Temperature = 20° C

Power = 300 watt

Wire's initial temperature = 20°C

Wire's final temperature = 1820°C

We need to calculate the resistance at 20°C

Using formula of power

$P=\dfrac{V^2}{R_{0}}$

$R_{0}=\dfrac{V^2}{P}$

Where, P = power

V = Potential difference

$R_{0}=\dfrac{150^2}{300}$

$R_{0}=75\ \Omega$

We need to calculate the resistance at 1820°C

Using formula of temperature of coefficient of resistivity

$R=R_{0}(1+\alpha(T-T_{0}))$

$R=75(1+0.0003125(1820-20))$

$R=117.19\ \Omega$

We need to calculate the power

Using formula of power

$P'=\dfrac{V^2}{R}$

$P'=\dfrac{150^2}{117.19}$

$P'=192\ Watt$

We need to calculate the percentage change in the power dissipation as the wire's temperature rises from 20°C to 1820°C

$\Delta P=|\dfrac{P'-P}{P}|\times100$

$\Delta P=|\dfrac{192-300}{300}|\times100$

$\Delta P=36\%$

Hence, The percentage change in the power dissipation as the wire's temperature rises from 20°C to 1820°C is 36%.

Answer 2

Answer:

The power decreases by 36%

Explanation:

Given:

At 20° C

Power, P₀ = 300 W

Potential difference, V = 150 volts

Now, power is given as

P = V²/R

where, R is the resistance

on substituting the values, we get

300 = 150²/R₀

or

R₀ = 75 Ω

Now, the variation of resistance with temperature is given as

R = R₀[1 + α(T - T₀)]

where, α is the temperature coefficient of resistivity = 0.0003125 (°C⁻¹)

now, at

T₀ = 20° C

R₀ = 75 Ω

for

T = 1820° C

we have

R = R₀[1 + α(T - T₀)]

substituting the values

we get

R = 75×[1 + 0.0003125 × (1820 - 20)]

or

R = 117.18 Ω

Now using the formula for power

We have,

P = V²/R

or

P = 150²/117.18 = 192 W

Therefore, the percentage change will be

= $\frac{P-P_0}{P_0}\times 100$

on substituting the values , we get

= $\frac{192-300}{300}\times 100$

= -36%

here, negative sign depicts the decrease in power