Question

Resistance of a coil is given as 15 $\pm 2$% ohm. When a voltage V is impressed on the resistor, the power dissipation P can be calculated in two different ways

$P=I^{2}R$ ... (iii)

$P=VI$ ... (iv)

Measurements were made as follows:

Supply voltage across the resistor V $=150\pm$1% and current flow through the resistor. $I=10\pm 1$% A. Then

• A. equation (iii) gives rise to smaller error and hence to be used for finding P.
• B. equation (iv) gives rise to smaller error and hence to be used for finding P.
• C. both equations (iii) and (iv) give same value of error.
• D. more data is necessary in order to say which equation is better (of iii and iv).

Answer 1

Answer: (B)

equation (iv) gives rise to smaller error and hence to be used for finding P.

If we use equation (iii)
$P=I^{2}R$
$\frac{\partial P}{\partial I}=2IR$
$=2\left(10\right)\left(15\right)$
$=300$
$\frac{\partial P}{\partial R}=I^2=100$

$\Delta P=\sqrt{{(\Delta I.\frac{\partial P}{\partial I})}^2+{(\Delta R.\frac{\partial P}{\partial R})}^2}$

$=\sqrt{(0.1\times 300{)}^2+(0.3\times 100{)}^2}$

$=\sqrt{(30{)}^2+(30{)}^2}=\sqrt{1800}$

$=42watt$

If we use equation (iv)
$P=VI$
$\Delta P=\sqrt{{(\Delta V.\frac{\partial P}{\partial V})}^2+{(\Delta I.\frac{\partial P}{\partial I})}^2}$

$=\sqrt{(1.5\times 10{)}^2+(0.1\times 150{)}^2}$

$=21watt$
$\Rightarrow$ equation (iv) gives smaller error and hence it is to be used.