Question

What is the equivalent resistance of the network of point between A and B (each of resistance value is r)​

Answer 1

Given :

Four resistors are connected as shown in the figure$.$

To Find :

We have to find equivalent resistance between A and B.

Concept :

★ Potential difference across each resistor remains same in parallel connection.

First of all we need to simplify the given circuit.

See the attachment for better understanding.

Calculation :

From the simplified circuit, It is clear that all four resistors are in parallel to each other.

⭆ 1/Req = 1/R₁ + ... + 1/R₄

⭆ 1/Req = 1/R + 1/R + 1/R + 1/R

⭆ 1/Req = 4/R

⭆ Req = R/4 Ω

Cheers!

Answer 2

Answer :

To Find :-

The Equivalent Resistance between A and B.

Given :-

• Value of each resistance = r

We know :-

⠀⠀⠀⠀⠀⠀Equivalent Resistance in parallel ⠀⠀⠀⠀⠀⠀circuit :-

$\boxed{\underline{\bf{\dfrac{1}{R_{P}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ... + \dfrac{1}{R_{n}}}}}$

Where :-

• $R_{p}$ = Equivalent Resistance in a Parallel Circuit.

• R = Resistance

Concept :-

Here the Resistance will be equal and it's value is r.

The equivalent Resistance here in this case is :

$\blue{\underline{\bf{\dfrac{1}{R_{P}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + \dfrac{1}{R_{4}}}}}$

Now , by using the Equation and substituting the value of Resistance in it , we can find the equivalent resistance in the circuit.

Solution :-

Given :-

• Resistance = r

Using the formula and substituting the values in it, we get :-

$:\implies \bf{\dfrac{1}{R_{P}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + \dfrac{1}{R_{4}}} \\ \\ \\ :\implies \bf{\dfrac{1}{R_{P}} = \dfrac{1}{r} + \dfrac{1}{r} + \dfrac{1}{r} + \dfrac{1}{r}} \\ \\ \\ :\implies \bf{\dfrac{1}{R_{P}} = \dfrac{1 + 1 + 1 + 1}{r}} \\ \\ \\ :\implies \bf{\dfrac{1}{R_{P}} = \dfrac{4}{r}} \\ \\ \\ :\implies \bf{R_{P} = \dfrac{r}{4}} \\ \\ \\ \therefore \purple{\bf{R_{P} = \dfrac{r}{4}\:\Omega}}$

Hence, the equivalent resistance between A and B is r/4 ohms.