Question

In an experiment value of two resistance are measured to be R1=(5.0+-0.2) ohm and R2=(10.0+-0.1) ohm. find the value of total resistance in (1) series, (2) parallel with limits of percentage error.

Answer 1

In an experiment value of two resistance are measured to be R1=(5.0+-0.2) ohm and R2=(10.0+-0.1) ohm. find the value of total resistance in (1) series, (2) parallel with limits of percentage error.

Answer 2

Answer:

The total resistance with limits of percentage error, when connected in series is $(15.0\pm0.5)\Omega$ and in parallel is $(3.3\pm0.7)\Omega$

Explanation:

Given data:

The resistance of resistor $R_{1} =(5.0\pm0.2)\Omega$

The resistance of resistor $R_{1} =(10.0\pm0.1)\Ohm$

(1) When the resistors are connected in series, the equivalent resistance is given by $R=R_{1} +R_{2}$

The total resistance of the given resistors connected in series is

$R=R_{1} +R_{2}=5+10=15\Omega$

Percentage error in series combination,

$\frac{\Delta R}{R} \times 100=\Big(\frac{\ 0.2}{5} +\frac{0.1 }{10}\Big)\times 100$

$=\big(0.04 +\ 0.01}\big)\times 100$

$=\big(0.05\big)\times 100=5\%$  

When connected in series, the total resistance with limits of percentage error is $(15.0\pm0.5)\Omega$.

(2) When the resistors are connected in parallel, the equivalent resistance is given by $\frac{1}{R} =\frac{1}{R_{1}} +\frac{1}{R_{2}}$

Then the total resistance of the given resistors connected in parallel is

${R} =\frac{R_{1}R_{2}}{R_{1}+R_{2} }$

$=\frac{5\times 10}{5+10}=\frac{50}{15} =3.3\Omega$

Percentage error in parallel combination,

$\frac{\Delta R}{R} \times 100=\Big(\frac{\ 0.2}{5} +\frac{0.1 }{10}+\frac{0.2+0.1 }{5+10}\Big)\times 100$

$=\big(0.04+0.01+0.02\big)\times 100$

$=\big(0.07\big)\times 100=7\%$  

When connected in parallel, he total resistance with limits of percentage error is $(3.3\pm0.7)\Omega$.