Question

AD RTD has ${\alpha }_0=0.005\left(l{/}^{\circ }C\right),R=500\Omega$ and a dissipation constant of $P_D=30\left(mW/°C\right)$ at 20°C. The RTD is used in a bridge circuit as shown in figure with $R_1=R_2=500\Omega$ and $R_3$ a variable resistor used to null the bridge. If the supply is 10V and the RTD is placed in a bath at 0°c, the value of $R_3$ (in $\Omega$ ) to null the bridge is

• A. 553
• B. 323
• C. 493
• D. 454

Answer 1

The correct option is D 454

The value of RTD resistance at $0^{\circ }C$ without including the effects of dissipation
$R=500[1+0.005(0-20\left)\right]=450\Omega$

If we exclude the effects of self-heating we would expect the bridge to null with $R_3$ equal to $450\Omega$. Now as we see the effect of self-heating for this problem, first we find the power dissipated in the RTD from the circuit
$P=I^{2}R$

$I=\frac{10}{500+450}=10.53mA$

$P=(10.53\times {10}^{-3}{)}^2\times 450$

$=49.9\left(mW\right)$
$\text{Temperature rise}\left(\Delta T\right)=\frac{P}{P_D}=\frac{49.9\left(mW\right)}{30\left(mW{/}^{\circ }C\right)}=1.66{(}^{\circ }C)$

Thus the RTD is not actually at bath temperature of $0^{\circ }C$ but at a temperatrue of ${1.66}^{\circ }C$

$R=500[1+0.005(1.66-20\left)\right]$
$=454.15\Omega$

Thus the bridge is null with $R=454\Omega$