Question

What is the resistance between P and Q?

Answer 1

Hint: We have to simplify the given circuit step by step by using the concept of parallel and series combinations of resistors. We observe the connection of resistors in the circuit and use the basic formula to simplify the circuit.

Formula used: 1. If two resistors $R_1$ and $R_2$ are connected in series, Then the equivalent resistance of these two resistors will be equal to $R_1+R_2$.
2. If two resistors $R_1$ and $R_2$ are connected in parallel, Then the equivalent resistance of these two resistors will be equal to $\dfrac{{R_1}\times {R_2}}{{R_1}+{R_2}}$.

Complete step by step answer:
To calculate the resistance between P and Q. First mark two points L and M at the joints in the given figure. It can be seen below:

In first step we will calculate the resistance across the branch LM where upper two resistance are in series with each other hence we can write it as,
${R_{ULM}} = 2\Omega + 2\Omega = 4\Omega $
Now adding the upper branch resistance with the third resistance present in the LM branch we will get,
${R_{LM}} = {R_{ULM}}||2\Omega $
We know, ${R_{||}} = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Now putting the values and using the above formula we will get,
${R_{LM}} = \dfrac{{4\Omega \times 2\Omega }}{{4\Omega + 2\Omega }} = \dfrac{8}{6}\Omega $
On simplifying we will get,
${R_{LM}} = \dfrac{4}{3}\Omega $

Now from the above figure we can see all the three resistance are in parallel with each other.
Hence the total resistance between P and Q is,
${R_{{\text{eq}}}} = 2\Omega + \dfrac{4}{3}\Omega + 2\Omega $
${R_{{\text{eq}}}} = \dfrac{{6\Omega + 4\Omega + 6\Omega }}{3} = \dfrac{{16}}{3}\Omega $
Therefore, the resistance between P and Q is $\dfrac{16}{3} \Omega$.

Note: Remember that the equivalent resistance is where the total resistance is connected either in parallel or in series. The electrical resistance shows how much energy will be required when we move the charges that are the current through the circuit.