Question

If two resistors of resistance R1=(4 ± 0.5)ohm and R2=(16 ± 0.5)ohm are connected (1) In series and (2) Parallel . Find the equivalent resistance in each case with limits of % error.

Answer 1

Hope this helps

Points to note

# Errors always add

# percentage error = relative error ×100

Answer 2

Given,

resistance of first resistor $R_1= 4$±$0.5$ ohm.

resistance of second resistor $R_2 = 16$±$0.5$ ohm.

total change in resistance for circuit is

$\Delta R = 0.5+0.5\\\Delta R = 1ohm$

case 1): resistors in series

equivalent resistance in series connection is given by

$R_{es}=R_1+R_2\\R_{es} = 4+16 \\R_{es} = 20 ohm$

here $R_{es}$ is series equivalent resistance.

Now, percentage error in series connection is

$\%e_s = \frac{\Delta R}{R_{es}}*100$

From the given values we know that change in resistance of two resistors will be

Then,

$\%e_s = \frac{1}{20}*100\\\\\%e_s = 0.05*100\\\%e_s = 5\%$

case 2): resistors in parallel

equivalent resistance in parallel connection is given by

$\frac{1}{R_{ep}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\ \frac{1}{R_{ep}}=\frac{1}{4}+\frac{1}{16} \\\\\frac{1}{R_{ep}} = \frac{5}{16}\\\\ R_{ep} = \frac{16}{5}ohm$

percentage error of parallel connection is

$\%e_p = \frac{\Delta R}{R_{ep}}*100 \\\\\%e_p= \frac{1}{\frac{16}{5} }*100 \\\\\%e_p = \frac{5}{16}*100\\\\\%e_p = 0.3125*100\\\%e_p = 31.25\%$

∴ case (1):

equivalent resistance in series $R_{es} = 20 ohm$

percentage error in series $\%e_s = 5\%$

 case (2):

equivalent resistance in parallel $R_{ep} = \frac{16}{5} ohms$

percentage error in parallel $\%e_p = 31.25\%$

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