Question

In the circuit shown in the figure, the current through-

A. The $3\Omega $ resistor is 1.00A
B. The $3\Omega $ resistor is 0.25A
C. The $4\Omega $ resistor is 0.50A
D. The $4\Omega $ resistor is 0.25A

Answer 1

Hint: We will find out the equivalent resistances one by one of each circuit by applying the formula for resistance in series and resistance in parallel. Then we will apply the formula of current i.e. $I = \dfrac{V}{R}$. Refer to the solution below for further doubts.

Formula used: $I = \dfrac{V}{R}$, ${R_{total}} = {R_1} + {R_2} + .....{R_n}$ and ${R_{total}} = \dfrac{1}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....\dfrac{1}{{{R_n}}}}}$

Complete Step-by-Step solution:
To find the current, let’s find the equivalent resistance first.
For series resistance-
$ \Rightarrow {R_{total}} = {R_1} + {R_2} + .....{R_n}$
For parallel resistance-
$ \Rightarrow {R_{total}} = \dfrac{1}{{\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....\dfrac{1}{{{R_n}}}}}$
Observe carefully that the resistance given in the last branch i.e. $2\Omega - 4\Omega - 2\Omega $ are in series. So, let the total resistance be R1 which will be-
$
   \Rightarrow {R_1} = 2 + 4 + 2 \\
    \\
   \Rightarrow {R_1} = 8\Omega \\
$
Now, this R1 is parallel to the other $8\Omega $. So, the equivalent resistance R2 will be-
$
   \Rightarrow {R_2} = \dfrac{1}{{\dfrac{1}{8} + \dfrac{1}{8}}} \\
    \\
   \Rightarrow {R_2} = \dfrac{1}{{\dfrac{{8 + 8}}{{8 \times 8}}}} \\
    \\
   \Rightarrow {R_2} = \dfrac{{8 \times 8}}{{8 + 8}} \\
    \\
   \Rightarrow {R_2} = 4\Omega \\
$
Now, as we can see that this R2 is in series with the $2\Omega $ and $2\Omega $. So, the equivalent resistance R3 will be-
$
   \Rightarrow {R_3} = 2 + 4 + 2 \\
    \\
   \Rightarrow {R_3} = 8\Omega \\
$
Now, again R3 is parallel to the next $8\Omega $ resistance. So, the resistance R4 will be-
$
   \Rightarrow {R_4} = \dfrac{1}{{\dfrac{1}{8} + \dfrac{1}{8}}} \\
    \\
   \Rightarrow {R_4} = \dfrac{1}{{\dfrac{{8 + 8}}{{8 \times 8}}}} \\
    \\
   \Rightarrow {R_4} = \dfrac{{8 \times 8}}{{8 + 8}} \\
    \\
   \Rightarrow {R_4} = 4\Omega \\
$
Now, this R4 is in series with the next $3\Omega $ and $2\Omega $. They are also forming the final equivalent resistance. So, the final equivalent resistance ${R_{eq}}$ will be-
$
   \Rightarrow {R_{eq}} = 3 + 4 + 2 \\
    \\
   \Rightarrow {R_{eq}} = 9\Omega \\
$
As we know that current is calculated by $I = \dfrac{V}{R}$
Putting the values in the above formula, we will have-
Where, $V = 9$ (given in the question) and $R = 9 = {R_{eq}}$
$
   \Rightarrow I = \dfrac{V}{{{R_{eq}}}} \\
    \\
   \Rightarrow I = \dfrac{9}{9} \\
    \\
   \Rightarrow I = 1A \\
$
Hence, option A is the correct option.

Note: Resistance is the restriction that a substance offers to the progression of electric current. It is spoken to by the uppercase letter R. The standard unit of resistance is the ohm, now and again worked out as a word, and some of the time represented by the uppercase Greek letter omega: $\Omega $.