# 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 $

Verified

119.1k+ views

**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.