MHB What is the required length of wire A?

[Drain Brain]

two wires A and B made from two different materials have temperature coefficient of resistance equal to 0.0025 and 0.0005 ohm per degree Celsius respectively. It is desired to make a coil of wire having a resistance of 1200 ohms with a temperature coefficient of 0.001, using a suitable length of the two given wires connected in series. Determine the required length of wire A. ?I know that I have to use equations that relates resistance with temperature

like

\frac{R1}{R2} = \frac{(T+t1)}{(T+t2)}

{\Delta}_{ t} = t2-t1

but they don't seem useful for me solve the problem.

please help with this problem. Any electrical engineers out there I need your help.

regards

 

[zzephod]

two wires A and B made from two different materials have temperature coefficient of resistance equal to 0.0025 and 0.0005 ohm per degree Celsius respectively. It is desired to make a coil of wire having a resistance of 1200 ohms with a temperature coefficient of 0.001, using a suitable length of the two given wires connected in series. Determine the required length of wire A. ?I know that I have to use equations that relates resistance with temperature

like

\frac{R1}{R2} = \frac{(T+t1)}{(T+t2)}

{\Delta}_{ t} = t2-t1

but they don't seem useful for me solve the problem.

please help with this problem. Any electrical engineers out there I need your help.

regards

You need at some point the resistance per unit length of the two wires.

Check units...

.

 

[Drain Brain]

You need at some point the resistance per unit length of the two wires.

Check units...

.

I still don't see it.

 

[zzephod]

I still don't see it.

There is not enough information in what you posted to solve this without some additional assumptions and or different units from those you have posted.

You have two unknowns the lengths $l_1$ and $l_2$ to solve for these you need to know the base resistances per unit length at the reference temperature $t_0$.

Also you need to know the difference between the reference temperature and the temperature at which you want the resistance to be 1200 Ohms.

.

 

[DavidCampen]

two wires A and B made from two different materials have temperature coefficient of resistance equal to 0.0025 and 0.0005 ohm per degree Celsius respectively. It is desired to make a coil of wire having a resistance of 1200 ohms with a temperature coefficient of 0.001, using a suitable length of the two given wires connected in series. Determine the required length of wire A. ?

To solve for the relative lengths of the 2 wires, assuming that the temperature coefficients are given as \[\frac{ohm}{(unit\,length)*(degree\,C)}\] (Also, I have multiplied temperature coefficients by 10,000 to minimize typing, this will not affect the answer.)
Then:

let X be the wire having TC=25 and x its length.
let Y be the wire having TC=5 and y its length.
Then to obtain a composite length of wire having TC = 10 we have:
\[\frac{25x + 5y}{x+y}=10\]
\[25x+5y = 10x+10y\]\[15x=5y\] \[3x=y\]
This gives you the necessary relative proportions of the 2 wires:
wire Y must be 3 times as long as wire X.

Now you need to know the resistance per unit length of these 2 wires so that you can calculate the necessary length of each to be combined to yield 1200 ohms.

 

[zzephod]

To solve for the relative lengths of the 2 wires, assuming that the temperature coefficients are given as \[\frac{ohm}{(unit\,length)*(degree\,C)}\] (Also, I have multiplied temperature coefficients by 10,000 to minimize typing, this will not affect the answer.)
Then:

let X be the wire having TC=25 and x its length.
let Y be the wire having TC=5 and y its length.
Then to obtain a composite length of wire having TC = 10 we have:
\[\frac{25x + 5y}{x+y}=10\]
\[25x+5y = 10x+10y\]\[15x=5y\] \[3x=y\]
This gives you the necessary relative proportions of the 2 wires:
wire Y must be 3 times as long as wire X.

Now you need to know the resistance per unit length of these 2 wires so that you can calculate the necessary length of each to be combined to yield 1200 ohms.

1. The normal definition of the temperature coefficient has different units, it has units of $K^{-1}$ not $\Omega K^{-1}$, that is it is the fractional change in resistance per Kelvin change in temperature.

2. If we assume that the units in the original question are wrong and the should just be the fractional change in resistance per Kelvin, and that we have specific resistances of the two wires of $r_1$ $\Omega m^{-1}$ and $r_2$ $\Omega m^{-1}$ at the reference temperature $T_0$ $K$, and lengths $l_1$ $m$ and $l_2$ $m$, we have the resistance of the two wires in series at temperature $T\ K$ is:

$$\begin{aligned}R(T)&=R_1(1+\alpha_1 \Delta T) + R_2 (1+\alpha_2 \Delta T)\\
&=R_1+R_2 + R_1\alpha_1 \Delta T + R_2 \alpha_2 \Delta T\\
&=(R_1+R_2)\left[1+\left(\frac{R_1 \alpha_1+R_2 \alpha_2}{R_1+R_2}\right)\Delta T\right]
\end{aligned}$$

where $R_1=r_1 l_1$, $R_2=r_2 l_2$, $\alpha_1$ and $\alpha_2$ are the two temperature coefficients, and $\Delta T=T-T_0$

So the temperature coefficient of the composite of the two wires in series is:

$$\alpha_3=\frac{R_1 \alpha_1+R_2 \alpha_2}{R_1+R_2} \ \ K^{-1}$$

But we require (at the base temperature presumably) that $R_1+R_2=r_1 l_1+r_2 l_2=1200 \ \Omega$, so we have:

$$\alpha_3=\frac{r_1 l_1 \alpha_1+r_2 l_2 \alpha_2}{1200} \ \ K^{-1}=0.001 \ K^{-1} $$

This can be simplified further to eliminate $r_1 l_1$ or $r_2 l_2$, which assuming the algebra is right gives:

$$\alpha_3=\alpha_1 + \frac{r_2 l_2(\alpha_2-\alpha_1)}{1200}=0.0025- \frac{0.002 r_2 l_2}{1200}=0.001$$

Which may be simplified to $R_2=r_2 l_2=900 \ \Omega$, and so $R_1=r_1 l_1=300 \ \Omega$.