Question

Which student measures the wrong equivalent resistance between A and B in the following given table:-

Student	Circuit	Equivalent resistance
Student A		\[1\Omega \]
Student B		$1.5\Omega $

Answer 1

Hint: We know that there are two possible combinations of resistors. One is the series combination and the other is a parallel combination. A series combination of resistors is the same in all the resistors and the potential difference of the combination is the sum of the potential difference of each resistor. While in parallel combination potential difference is the same for each resistor but the current for the system is the sum of currents of each resistor.

Complete step-by-step solution
We know that In series connection equivalent resistance is the sum of all the individual resistances and in parallel connection the reciprocal of equivalent resistance is the sum of reciprocal of the individual resistances.
For student A, we have the value of each resistor equal to $1\Omega $. Two resistors (${{R}_{1}}$and ${{R}_{2}}$) are in series and the other one(${{R}_{3}}$) in parallel with them.

Resistance for the two resistors in series will be given as${{R}_{s}}={{R}_{1}}+{{R}_{2}}$. Therefore, it will be
${{R}_{s}}=1+1=2\Omega $.
Now, this ${{R}_{s}}=2\Omega $is in parallel with the third one
Hence using $\dfrac{1}{{{R}_{p}}}=\dfrac{1}{{{R}_{s}}}+\dfrac{1}{{{R}_{3}}}$we get
$\Rightarrow \dfrac{1}{{{R}_{p}}}=\dfrac{1}{{{R}_{s}}}+\dfrac{1}{{{R}_{3}}}$
$\Rightarrow \dfrac{1}{{{R}_{p}}}=\dfrac{1}{2}+\dfrac{1}{1}$
$\Rightarrow \dfrac{1}{{{R}_{p}}}=\dfrac{1+2}{2}$
$\Rightarrow \dfrac{1}{{{R}_{p}}}=\dfrac{3}{2}$
Therefore, ${{R}_{p}}=\dfrac{2}{3}\Omega $
Hence, equivalent resistance is $0.67\Omega $
Therefore, Student A is incorrect.
For student B, we have ${{R}_{2}}$and ${{R}_{3}}$in parallel with each other and ${{R}_{1}}$in series with them.

Therefore, equivalent resistance ${{R}_{eq}}$is given as follows:-
${{R}_{eq}}={{R}_{1}}+ {{R}_{s}}$
where,
$\left( \dfrac{1}{{{R}_{s}}} \right)= \left( \dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}} \right)$
$\Rightarrow \left( \dfrac{1}{{{R}_{s}}} \right) = \left( \dfrac{1}{1}+\dfrac{1}{1} \right)$
$\Rightarrow \left( \dfrac{1}{{{R}_{s}}} \right) = \left( \dfrac{1+1}{1} \right)$
$\Rightarrow \left( \dfrac{1}{{{R}_{s}}} \right) = \left( \dfrac{2}{1} \right)$
Using reciprocal rule for effective resistance in parallel connection we get
$\Rightarrow {{R}_{eq}}=1+\left( \dfrac{1}{2} \right)$
$\Rightarrow {{R}_{eq}}=\dfrac{2+1}{2}$
$\Rightarrow {{R}_{eq}}=\dfrac{3}{2}$
$\Rightarrow {{R}_{eq}}=1.5\Omega $
Hence, student B is correct.
Thus, we get the final answer, student A is incorrect and student B is correct.

Note: In solving the problems related to equivalent resistance we have to identify the correct arrangements of resistances. We should not get confused between the formulae of series and parallel combinations. In a parallel combination, we have to find the reciprocal of each resistance.