Question

What is the resultant resistance of the given circuit?

A) $20\Omega $
B) $8\Omega $
C) $7.2\Omega $
D) $12\Omega $

Answer 1

Hint: To solve this circuit we can clearly see that $8\Omega $ resistance and $12\Omega $ resistance are connected in parallel. First we find the resultant resistance of these two resistances by applying Formula of equivalent resistance for parallel combination.
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....... + \dfrac{1}{{{R_n}}}$
Where ${R_{eq}}$ is the equivalent or resultant resistance of n parallel resistance.
And $7.2\Omega $ resistance connected to these resistances in series combination apply here formula for series combination.
${R_{eq}} = {R_1} + {R_2} + ...... + {R_n}$
Where ${R_{eq}}$ is the resultant resistance of n series connected resistance.

Complete step by step answer:
First we should understand what is the resultant or equivalent resistance of any circuit.
The equivalent resistance of a network is that single resistor that could replace the entire network in such a way that for a certain applied voltage V you get the same current I as you were getting for a network.
Step 1

Here we see in the circuit diagram 8 ohm resistance and 12 ohm resistance are connected in parallel combination here a question arise what is the parallel combination Answer is, in the parallel combination first terminal of every resistance connected together and the second terminal of all resistances connected together
As given in question for $8\Omega $ and $12\Omega $ resistance.
First we solve this part of circuit
                   

Apply formula for parallel combination
$ \Rightarrow \dfrac{1}{{{R_{12}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
Where $
  {R_1} = 8\Omega \\
  {R_2} = 12\Omega \\
 $
And ${R_{12}}$ we take as resultant of ${R_1}$ and ${R_2}$
Put the values and solve
$ \Rightarrow \dfrac{1}{{{R_{12}}}} = \dfrac{1}{8} + \dfrac{1}{{12}}$
$ \Rightarrow \dfrac{1}{{{R_{12}}}} = \dfrac{{3 + 2}}{{24}}$
$ \Rightarrow \dfrac{1}{{{R_{12}}}} = \dfrac{5}{{24}}$
Now inverting this fraction
$ \Rightarrow {R_{12}} = \dfrac{{24}}{5} = 4.8\Omega $
We get the resultant of resistance of 8 ohm resistance and 12 ohm resistance is ${R_{12}} = 4.8\Omega $
Step 2
Now we can replace ${R_1}$ and ${R_2}$ by ${R_{12}}$ in the circuit then circuit become

Now we see that the resistance ${R_3}$ and the resistance ${R_{12}}$ are connected in series
And now what is the series connection the answer is in the series connection first terminal of first resistance connected to the battery and the second terminal of same resistance connected to the first terminal of the next resistance and so on in the last the 2nd terminal of the last resistance connected to the second terminal of battery as shown in figure,

We have formula for series connection
$ \Rightarrow {R_{eq}} = {R_3} + {R_{12}}$
Put the value of ${R_3}$ and ${R_{12}}$
$
   \Rightarrow {R_{eq}} = 7.2 + 4.8 \\
   \Rightarrow {R_{eq}} = 12\Omega \\
 $
And finally we get the equivalent resistance off ${R_3}$ and ${R_{12}}$
$\therefore {R_{eq}} = 12\Omega $
Hence the equivalent resistance of entire circuit is
$ \Rightarrow 12\Omega $
 $\therefore $ In this question option D is correct.

Note: There is another method to identify the combination of resistance in parallel combination the voltage across the all resistance is same and in the series combination the current should be same in all resistance why this we can also identify the combination of resistance and we can easily solve these types of question by applying these two formula end a common mistake done by the students is in the parallel combination They forgot to inverse the Fraction just applying the formula for parallel combination we should remember this thing.