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The effective resistance between the points \$A\$ and \$B\$ in the circuit shown in Fig. will be:(A) \$6\Omega \$(B) \$3\Omega \$(C) \$15\Omega \$(D) \$10\Omega \$

Last updated date: 25th Jun 2024
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Hint: We will use the general formula of adding resistances here. The net resistance between points \$A\$ and \$B\$ is the same as the addition of net resistances between points \$A\$ and \$C\$, \$C\$ and \$D\$, \$D\$ and \$B\$ which is given in the below figure. We will simplify the resistances using the formulas below.
Formula used: Resistance calculation in series \${R_{series}} = {R_1} + {R_2} + ...\$
Resistance calculation in parallel \$\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ...\$

Complete Step by step solution:

In the given diagram,
Net resistance between \$A\$ and \$B\$ is calculated by adding the net resistances between \$A\$ and \$C\$, \$C\$ and \$D\$, \$D\$ and \$B\$ since they will be in series.
Therefore, \${R_{AC}} = 1\Omega \$ as can be seen, from the figure.
In the calculation of \${R_{CD}}\$, we see that the three \$1\Omega \$ resistances are in series. We use the series resistance formula here to get the net resistance as \$1\Omega + 1\Omega + 1\Omega = 3\Omega \$ on the top branch.
Similarly, the three \$2\Omega \$ resistances are also in series. We use the series resistance formula again to get the net resistance as \$2\Omega + 2\Omega + 2\Omega = 6\Omega \$ on the bottom branch.
These \$3\Omega \$ net resistance of the top branch, \$2\Omega \$ resistance of the middle branch, and \$6\Omega \$ net resistance of the bottom branch are now in parallel.
Hence, we use the formula for resistances in parallel to get the net resistance as \$\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{3\Omega }} + \dfrac{1}{{2\Omega }} + \dfrac{1}{{6\Omega }}\$
\$ \Rightarrow \dfrac{1}{{{R_{parallel}}}} = \dfrac{{2 + 3 + 1}}{{6\Omega }} = \dfrac{1}{{1\Omega }}\$
\$\therefore {R_{parallel}} = 1\Omega \$
This is the net resistance between points \$C\$ and \$D\$. Thus \${R_{CD}} = 1\Omega \$.
The resistance between points \$D\$ and \$B\$, i.e. \${R_{DB}} = 1\Omega \$ as can be seen from the figure given.
Now the resistances \${R_{AC}}\$, \${R_{CD}}\$ and \${R_{DB}}\$ are in series.
We will therefore use the series formula for summation of resistances.
\${R_{net}} = {R_{AC}} + {R_{CD}} + {R_{DB}} = 1\Omega + 1\Omega + 1\Omega = 3\Omega \$
\$ \Rightarrow {R_{net}} = 3\Omega \$
Thus the net resistance between the points \$A\$ and \$B\$ is equal to \$3\Omega \$.

Therefore option (B) is the correct answer.

Note: The other process of calculating the equivalent or net resistance between two points includes assuming a current of \$1A\$ flowing through the circuit and calculating the net voltage drops from a specific point. After finding the net voltage drop, we will use the formula \$V = IR\$, to find the \$R\$, i.e. effective resistance in case of complicated circuits.