Question

In an experiment, the values of two resistances were measured to be as given below: 

R1 =5+0.2 ohm; R2 = 10+0.1 ohm 

Find the values of total resistance in (i) series and (ii) parallel with limits of possible percentage error in each case​

Answer 1

Answer:

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Answer 2

Answer:

The total resistance with percentage error, in series combination is $(3.3+7)\Omega$ and in parallel combination is $(3.3+7)\Omega.$

Explanation:  

Given the resistance of resistor,  $R_{1} =5+0.2 \Omega$

The resistance of other resistor,  $R_{1} =10+0.1 \Omega$

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

The total resistance of the resistors connected in series is

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

And the percentage error in series combination,

$\frac{\Delta R}{R} \times 100=\frac{\Delta R_{1} }{R_{1}} \times 100+\frac{\Delta R_{2} }{R_{2}} \times 100$

$\implies \frac{\Delta R}{R} \times 100=\Big(\frac{\Delta R_{1} }{R_{1}}+\frac{\Delta R_{2} }{R_{2}}\Big) \times 100$

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

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

$=0.05 \times 100$

$=5 \%$

Therefore, when the resistors are connected in series, the total resistance with limits of percentage error is $(15+5)\Omega$.

(2) When the resistors are connected in parallel, the equivalent resistance of the combination is given by

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

Then the total resistance of the resistors connected in parallel is

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

$=\frac{5\times 10}{5+10}$

$=\frac{50}{15}=3.3\Omega$

And the percentage error in parallel combination,

$\frac{\Delta R}{R} \times 100=\Big(\frac{\Delta R_{1} }{R_{1}}+\frac{\Delta R_{2} }{R_{2}}+\frac{\Delta R_{1}+ \Delta R_{2}}{R_{1}+R_{2}}\Big) \times 100$

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

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

$=0.07 \times 100=7\%$

Therefore, when the resistors are connected in parallel, the total resistance with limits of percentage error is $(3.3+7)\Omega$