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The effective resistance between P and Q for the following network is:

A) $\dfrac{1}{{12}} \Omega$
B) $21\Omega $
C) $12\Omega $
D) $\dfrac{1}{{21}}\Omega $

seo-qna
Last updated date: 24th Feb 2024
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IVSAT 2024
Answer
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Hint: In the given circuit we can see that many resistances are connected. In circuits resistances can be connected in series and parallel combinations. The net resistance between points P and Q can be found by solving these combinations. So we are going to use the following formulae of series and parallel combinations of resistances.
Series combination:
\[R = {R_1} + {R_2}\]
Parallel combination:
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$

Complete step by step solution:
Let $4\Omega $resistance is ${R_1}$, $3\Omega $ resistance is ${R_2}$, other $3\Omega $ resistance is ${R_3}$, $5\Omega$ resistance is ${R_4}$ and $6\Omega$ resistance is ${R_5}$.
Between points A and B resistances \[{R_2}\] and \[{R_3}\] are in series.
Let their resultant is \[{R'}\].
So \[{R'} = {R_2} + {R_3}\]
Putting the values of \[{R_2}{\text{ and }}{R_3}\]
\[\Rightarrow {R'} = 3 + 3\]
\[\Rightarrow {R'} = 6{\text{ }}\Omega \]
\[\Rightarrow {R'}\] and \[{R_5}\] are in parallel. Let their resultant is \[{R^{''}}\].
$\Rightarrow \dfrac{1}{{{R^{''}}}} = \dfrac{1}{{{R'}}} + \dfrac{1}{{{R_5}}}$
Putting the values of \[{R'}\] and \[{R_5}\]
$\Rightarrow \dfrac{1}{{{R^{''}}}} = \dfrac{1}{6} + \dfrac{1}{6}$
$\Rightarrow \dfrac{1}{{{R^{''}}}} = \dfrac{2}{6}$
$\Rightarrow \dfrac{1}{{{R^{''}}}} = \dfrac{1}{3}$
$\Rightarrow {R^{''}} = 3{\text{ }}\Omega $
Now,\[{R^{''}}\], \[{R_1}\]and \[{R_4}\]are in series. Let their resultant is R.
So, \[R = {R^{''}} + {R_1} + {R_4}\]
Putting the values of \[{R^{''}},{R_1}{\text{ and }}{R_4}\]
$\Rightarrow R = 4 + 3 + 5$
$\Rightarrow R = 12\Omega $

The resultant resistance between points P and Q is $R = 12 \Omega $.

Note: In the questions where circuits are presented we have to be careful while finding the series and parallel combinations of resistances. The circuit given in the question has resistances only so it is sort of basic and easy. But circuits can be very complicated when capacitors and inductors are also involved in the circuit. In such conditions we have to find the total resistance considering the impedance created by the capacitors and inductors also. The impedance in case of capacitor is given by following formula,
${X_c} = \dfrac{1}{{\omega C}}$
Where, $\omega $ is the angular frequency ${s^{ - 1}}$ and
C is the capacitance in coulomb
The impedance in case of inductor is given by following formula,
${X_c} = \omega L$
Where, $\omega $ is the angular frequency ${s^{ - 1}}$ and
L is the inductance in Henry
So circuits consisting of capacitors, inductors we will use above formulae to find net impedance.