Question

Show how would you join three resistors, each of resistance $9\,\Omega$ so that the equivalent resistance of the combination is $13.5\,\Omega$?

Answer 1

From the diagram, it is clear that the two resistors are in parallel with the one resistor in the series. First calculate the equivalent resistance of the two parallel resistors using formula. Then with the equivalent resistance calculated, again find the total resistance with the series resistors using the formula.

Useful formula:
(1) The equivalent resistance of the resistors connected in parallel is given by
$r = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Where $r$ is the equivalent resistor, ${R_1}$ is the first resistor and ${R_2}$ is the second resistor.
(2) The equivalent resistance of the resistors that connected in series,
${R_e} = r + R$
Where ${R_e}$ is the equivalent resistance.

Complete step by step solution:
It is given that the
Resistance of each resistor, ${R_1} = {R_2} = {R_3} = 9\,\Omega$
The equivalent resistance of the combination, $r = 13.5\,\Omega$
The formula of the equivalent resistance is taken.
$r = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Substituting the values in the above step.
$r = \dfrac{{9 \times 9}}{{9 + 9}}$
By performing simple arithmetic operations.
$r = 4.5\,\Omega$
Using the formula (2) to calculate the equivalent resistance.
${R_e} = r + R$
Substituting the values in the above formula,
${R_e} = 4.5 + 9$
${R_e} = 13.5\,\Omega$

Hence the equivalent resistance of the given circuit is obtained as $13.5\,\Omega$.

Note: Normally adding the resistance in the circuit gives the equivalent resistance, if they are connected in series. Remember the formula of the equivalent resistance in parallel. The parallel circuit has the same voltage in each resistor and the series circuit has the same current in each circuit.