Question

A Wheatstone bridge has the resistance $10\,\Omega$, $10\,\Omega$, $10\,\Omega$ and $30\,\Omega$ in its four arms. What resistance joined in parallel to $30\,\Omega$ resistance will bring it to the balanced condition?
A. $2\,\Omega$
B. $5\,\Omega$
C. $10\,\Omega$
D. $15\,\Omega$

Answer 1

Use the equation for the equivalent resistance of two resistors connected in parallel. Also, use the conditions that will bring the Wheatstone’s bridge in the balanced condition. Determine the equivalent resistance of the two resistors connected in parallel as given in the question and substitute it in the balanced condition of Wheatstone’s bridge.

Formulae used:
The equivalent resistance ${R_{eq}}$ of the two resistance connected in parallel is given by
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ …… (1)
Here, ${R_1}$ is the resistance of the first resistor and ${R_2}$ is the resistance of the second resistor.

Complete step by step answer:
We have given that the four resistors with resistances $10\,\Omega$, $10\,\Omega$, $10\,\Omega$ and $30\,\Omega$ are connected in four arms of a Wheatstone’s bridge.
${R_1} = 10\,\Omega$, ${R_2} = 10\,\Omega$, ${R_3} = 10\,\Omega$, ${R_4} = 30\,\Omega$
We know that if four resistors with resistances ${R_1}$, ${R_2}$, ${R_3}$ and ${R_4}$ are connected in a Wheatstone’s bridge, the balanced condition for the Wheatstone’s bridge is
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{R_3}}}{{{R_4}}}$ …… (2)
We need to connect a resistor with resistance $R$ in parallel to the resistor ${R_4}$.
The equivalent resistance of the two resistances $R$ and ${R_4}$ can be determined by using equation (1).
Substitute $R$ for ${R_1}$ and ${R_4}$ for ${R_2}$ in equation (1).
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{R} + \dfrac{1}{{{R_4}}}$
Substitute $30\,\Omega$ for ${R_4}$ in the above equation.
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{R} + \dfrac{1}{{30\,\Omega }}$
$\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{{30 + R}}{{30R}}$
$\Rightarrow {R_{eq}} = \dfrac{{30R}}{{30 + R}}$
The condition for the given Wheatstone’s bridge to be in balanced condition shown in equation (2) becomes
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{R_3}}}{{{R_{eq}}}}$
Substitute $10\,\Omega$ for ${R_1}$, ${R_2}$ and ${R_3}$ and $\dfrac{{30R}}{{30 + R}}$ for ${R_{eq}}$ in the above equation and solve the above equation for $R$.
$\dfrac{{10\,\Omega }}{{10\,\Omega }} = \dfrac{{10\,\Omega }}{{\dfrac{{30R}}{{30 + R}}}}$
$\Rightarrow 1 = \dfrac{{10\,\Omega }}{{\dfrac{{30R}}{{30 + R}}}}$
$\Rightarrow 30R = 10\left( {30 + R} \right)$
$\Rightarrow 30R = 300 + 10R$
$\Rightarrow 30R - 10R = 300$
$\Rightarrow 20R = 300$
$\Rightarrow R = \dfrac{{300}}{{20}}$
$\Rightarrow R = 15\,\Omega$

Therefore, the resistance joined in parallel that will bring the balanced condition is $15\,\Omega$.

Hence, the correct option is D.

Note:
The students should use the balanced condition for the Wheatstone’s bridge carefully. If the condition for the Wheatstone’s bridge is not used correctly, the final answer we got by the calculations is not the correct. Also don’t forget to use the equivalent resistance of the two resistors connected in parallel.