Question

Two resistors of resistances ${R_1} = 100 \pm 3\;{\rm{ohm}}$ and ${R_2} = 200 \pm 4\;{\rm{ohm}}$ are connected (a) in series (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for
(a) the relation $R = {R_1} + {R_2}$ and for
(b) $\dfrac{1}{{R'}} = \dfrac{1}{{{R_{\kern 1pt} }_1}} + \dfrac{1}{{{R_2}}}$ and $\dfrac{{\Delta R'}}{{R{'^2}}} = \dfrac{{\Delta {R_1}}}{{R_1^2}} + \dfrac{{\Delta {R_2}}}{{R_2^2}}$.

Answer 1

Hint: First, identify the true values and the errors in the given resistances. Then, substitute the values in the equations given in the question to obtain the values of unknown quantities. The final resistance will contain the true value term and the error term.

Complete step by step answer:
The resistance of the first resistor, ${R_1} = 100 \pm 3\;\Omega $.
The resistance of the second resistor, ${R_2} = 200 \pm 4\;\Omega $.

(a)
The equivalent resistance of the series combination is given by,
$R = {R_1} + {R_2}$
Now, we will substitute the values of ${R_1}$ and ${R_2}$ in the above equation to find the equivalent resistance.
$
R = \left( {100 \pm 3\;\Omega } \right) + \left( {200 \pm 4\;\Omega } \right)\\
 = \left( {100 + 200} \right) \pm \left( {3 + 4} \right)\;\Omega \\
 = 300 \pm 7\;\Omega
$
Therefore, the equivalent resistance of the series combination is $300 \pm 7\;\Omega $.

(b)
The expression to find the equivalent resistance of the parallel combination without error limit is given by,
$\dfrac{1}{{R'}} = \dfrac{1}{{{R_{\kern 1pt} }_1}} + \dfrac{1}{{{R_2}}}$
Here, $R'$ is the equivalent resistance of the parallel combination without error limit.
We will rewrite the above equation as,
$R' = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$
The above equation will give the equivalent resistance without the error limits. Hence, we will use only the true values of the resistances in the equation.
Substituting $100\;\Omega $ for ${R_1}$ and $200\;\Omega $ for${R_2}$in the above equation, we get
$
R' = \dfrac{{100\;\Omega \times 200\;\Omega }}{{100\;\Omega + 200\;\Omega }}\\
 = \dfrac{{20000\;\Omega }}{{300\,\Omega }}\\
 \cong 66.7\;\Omega
$
The relation to find the error in the equivalent resistance of the parallel combination is given in the question as,
$\dfrac{{\Delta R'}}{{R{'^2}}} = \dfrac{{\Delta {R_1}}}{{R_1^2}} + \dfrac{{\Delta {R_2}}}{{R_2^2}}$
Here, $\Delta R'$ is the error in $R'$, $\Delta {R_1}$ is the error in ${R_1}$ and $\Delta {R_2}$ is the error in ${R_2}$.
The value with the $ \pm $ symbol in front of it in the resistance value is the error in the resistance.
Hence, we substitute $3\;\Omega $ for $\Delta {R_1}$, $4\;\Omega $ for $\Delta {R_2}$, $100\;\Omega $ for ${R_1}$ and $200\;\Omega $ for${R_2}$ and $66.7\;\Omega $ for $R'$ in the above equation to find the error in $R'$.
$
\dfrac{{\Delta R'}}{{R{'^2}}} = \dfrac{{\Delta {R_1}}}{{R_1^2}} + \dfrac{{\Delta {R_2}}}{{R_2^2}}\\
\dfrac{{\Delta R'}}{{{{\left( {66.7\;\Omega } \right)}^2}}} = \dfrac{{3\;\Omega }}{{{{\left( {100\;\Omega } \right)}^2}}} + \dfrac{{4\;\Omega }}{{{{\left( {200\;\Omega } \right)}^2}}}\\
\Delta R' = 3\;\Omega {\left( {\dfrac{{66.7\;\Omega \;}}{{100}}} \right)^2} + 4\;\Omega {\left( {\dfrac{{66.7\;\Omega }}{{200}}} \right)^2}\\
 \cong 1.8\;\Omega
$
Hence, we write the equivalent resistance with the error limit as
$R' \pm \Delta R' = 66.7 + 1.8\;\Omega $
Therefore, the value of the equivalent resistance of the parallel combination is $66.7 + 1.8\;\Omega $.

Note:
It should be noted that the errors and the true values of the resistances should be added separately to find the resistance of the series combination. All the operations such as addition, subtraction, multiplication and addition should be done separately for true values and the errors in any measurements.