Question

Two resistors of resistances ${R_1} = 100 \pm 3$ ohm and ${R_2} = 200 \pm 4$ 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) $\dfrac{1}{{R\prime }} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ and $\dfrac{{\Delta R^\prime }}{{{R^{{\prime ^2}}}}} = \dfrac{{\Delta {R_1}}}{{R_1^2}} + \dfrac{{\Delta {R_2}}}{{R_2^2}}$

Answer 1

Hint: First we need to find the equivalent resistance for both the connections that are series and parallel connection. Now according to the question using the relations given for solving the questions to get the proper answer.

Complete step-by-step answer:
(a)We know that if we need to find the equivalent resistance for two or more resistors in series then,
${R_{eqs}} = {R_1} + {R_2}$
Therefore,
${R_{eqs}} = (100 \pm 3) + (200 \pm 4)\Omega$
${R_{eqs}} = (300 \pm 7)\Omega$, this is the equivalent resistance for the two resistors in series.
(b) We know that for two or more resistance connected parallel to each other then the equivalent resistance is,
${R_{eqp}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$ ,
 ${R_{eqp}} = \dfrac{{100 \times 200}}{{100 + 200}}$,
${R_{eqp}} = \dfrac{{200}}{3}$,
${R_{eqp}} = 66.7\Omega$, this is the equivalent resistance when the two resistors are connected in parallel.

Now taking the second equation,
$\dfrac{{\Delta R^\prime }}{{{R^{{\prime ^2}}}}} = \dfrac{{\Delta {R_1}}}{{R_1^2}} + \dfrac{{\Delta {R_2}}}{{R_2^2}}$,
On simplifying this equation,
\[\Delta R\prime = {R^{{\prime ^2}}}\left( {\dfrac{{\Delta {R_1}}}{{R_1^2}} + \dfrac{{\Delta {R_2}}}{{R_2^2}}} \right)\]
\[\Delta R\prime = {\left( {\dfrac{{200}}{3}} \right)^2}\left( {\dfrac{3}{{{{100}^2}}} + \dfrac{4}{{{{200}^2}}}} \right)\],
On solving the above equation we get,

\[\Delta R\prime = \left( {\dfrac{4}{9}} \right)\left( {\dfrac{3}{1} + \dfrac{4}{4}} \right)\],
\[\Delta R\prime = \left( {\dfrac{{16}}{9}} \right)\],
\[\Delta R\prime = 1.8\Omega\]

Therefore,
\[R\prime = \left( {66.7 \pm 1.8} \right)\Omega\]

Additional Information:
Resistance is the opposition to the flow to current. It opposes the free flow of electrons making the electrons flow slowly thus current decreases.
In series combination the greater the resistance the greater is the equivalent resistance, but in the case of a parallel resistor circuit the greater the resistance the lower the equivalent resistance gets.

According to ohm’s law, we know that V=IR, where V is the potential difference or the voltage, I is the current, and R is the resistance applied in the circuit. From this relation, we can say that the potential difference or voltage V is directly proportional to the current(I) and resistance(R).
Resistance is calculated in ohm.
Current in ampere.
Voltage in volts.

Note: Find the solution for both the series and parallel circuit. In calculation of equivalent resistance for series circuit ${R_{eq}} = {R_1} + {R_2}$ and for parallel circuit we can use ${R_{eq}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$ or \[{R_{eq}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\]