Question

Two resistances connected in parallel give the resultant value of 2 ohms, when connected in series the value becomes 9 ohms. Calculate the value of each resistance.

Answer 1

Hint: The resistance of the wire opposes the flow of electrons, so is the flow of current. The voltage drop across the resistors remains the same in the parallel combination of the resistors. The current through each resistor remains the same in the series combination of the resistors.

Complete step by step answer:
Given: The resultant resistance of parallel combination is ${R_p} = 2\;\Omega $, the resultant resistance of series combination is ${R_s} = 9\;\Omega $.
Use the expression of the resultant resistance of parallel combination to find the resultant value of the resistance. The expression is given as,
${R_p} = \dfrac{{{R_1} \cdot {R_2}}}{{{R_1} + {R_2}}}......\left( 1 \right)$
Here, ${R_1}$ and ${R_2}$ are the value of each resistance.
Use the expression of the resultant resistance of series combination to find the resultant value of the resistance. The expression is given as,
${R_s} = {R_1} + {R_2}......\left( 2 \right)$
Substitute ${R_s}$ for ${R_1} + {R_2}$ in the equation (1).
${R_p} = \dfrac{{{R_1} \cdot {R_2}}}{{{R_s}}}......\left( 3 \right)$
Substitute $9\;\Omega $ for ${R_s}$ and $2\;\Omega $ for ${R_p}$ in the equation (3).
$\Rightarrow 2\;\Omega = \dfrac{{{R_1} \cdot {R_2}}}{{9\;\Omega }}$
On simplification,
$\Rightarrow {R_1} = \dfrac{{18\;{\Omega ^2}}}{{{R_2}}}........\left( 4 \right)$
Substitute \[\dfrac{{18\;\Omega }}{{{R_2}}}\] for ${R_1}$ and $9\;\Omega $ for ${R_s}$in the equation (2).
$\Rightarrow 9\;\Omega = \dfrac{{18\;\Omega }}{{{R_2}}} + {R_2}$
$\Rightarrow R_2^2 - 9{R_2} + 18\;\Omega = 0$
On simplification,
$\Rightarrow \left( {{R_2} - 3\;\Omega } \right)\left( {{R_2} - 6\;\Omega } \right) = 0$
$\Rightarrow {R_2} = 3\;\Omega ,6\;\Omega $
Substitute $3\;\Omega $ for ${R_2}$ in the equation (4) to find ${R_1}$.
$\Rightarrow {R_1} = \dfrac{{18\;{\Omega ^2}}}{{3\;\Omega }}$
On simplification,
$\Rightarrow {R_1} = 6\;\Omega $
Substitute $6\;\Omega $ for ${R_2}$ in the equation (4) to find ${R_1}$.
$\Rightarrow {R_1} = \dfrac{{18\;{\Omega ^2}}}{{6\;\Omega }}$
On simplification,
$\Rightarrow {R_1} = 3\;\Omega $

Therefore, the value of each resistance is $6\,\Omega ,3\;\Omega $ or $3\,\Omega ,6\;\Omega $.

Note:
- The resistance depends on the length and area of the wire. The parallel combination of electrical devices is used for domestic wiring.
- Use the expression for parallel combination and series combination carefully and remember always that the voltage drops across each resistance in parallel combination and current through each resistance in series combination remain the same.