Question

The resultant value of resistances each of value n ohms when connected in parallel is x , when these resistances are connected in series the resultant resistance is :​

Answer 1

Answer:

Given:

When n resistors are connected in Parallel, the net resistance is x

To find:

Net resistance when these resistors are connected in series.

Calculation :

Let resistance of each resistor be r

So in parallel:

$net \: resistance = \dfrac{r}{n} = x$

$= > r = x \times n$

Now in series combinations:

$net \: r = r + r + r....n \: times$

$= > net \: r = n \times r$

$= > net \: r = (x \times n) \times n$

$= > net \: r = x \times {n}^{2}$

So final answer is :

$\boxed{\boxed{ \red{eq. \: resistance = x {(n)}^{2}}}}$

Answer 2

Question :-

→ The resultant value of resistance each of value n $\Omega$ .when connected in parallel is x , when these resistances are connected in series the resultant resistance is -

Answer :-

$\to \boxed{ \sf r_{n} \: = x \: {n}^{2} } \\ \:$

To Find :-

→ The resultant resistance when these are in series combination .

Solution :-

According to the question,

→ When resistors are connected in parallel resultant resistance is x and each has value n $\Omega$ .

Let, resistance of each resistor is r ohm

hence , according to the question

$\to \sf \: total \: resistance \: =x \\ \\ \to \sf \frac{r}{n} = x \\ \: \\ \therefore \boxed{\sf \: r \: = n \: x} \: \:$

When resistors are in series combination ,then resultant resistance is ,

$\mapsto \boxed{\sf \: total \: resistance \: ( r_{n})\: = \sum_{i = 1}^{n} \:r_{i} }\: \\ \\ \sf all \: have \: same \: value \: r \\ \\ \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf\because \: r = nx } \\ \: \\ \to \sf \: r_{n} \: = (x \times n) \: \times n(times) \\ \\ \to \boxed{ \sf r_{n} \: = x \: {n}^{2} }$

Hence net resistance when these are in series is

$\: \sf r_{n} \: = x \: {n}^{2}$