Question

A wire of resistance of (r) is cut into 5 equal pieces. These pieces are connected in parallel with the resulted resistance of this combination is r1, then the relation R/r1 is____???​

Answer 1

Answer:

Explanation:

Given :-

Resistance of each wire = $\sf \frac{R}{5}$

To Find :-

$\sf \frac{R}{R_1} =??$

Solution :-

When they are connected in parallel, equivalent resistance will be calculated as

$\sf\implies \frac{R}{R_1} =\frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}}$

$\sf\implies \frac{R}{R_1} =\frac{5}{\frac{R}{5} }$

$\sf\implies \frac{R}{R_1} =\frac{25}{R}$

$\bf\implies \frac{R}{R_1} =25$

Hence, the relation R/r1 is 25.

Answer 2

$\bold{\underline{\underline{\sf{Answer:}}}}$

$\bold{\dfrac{R}{r_1}}$ = 25

$\bold{\underline{\underline{\sf{Step\:-\:by\:-\:step\:explanation:}}}}$

Given :

• A wire of resistance of (r)
• Wire is cut into 5 equal pieces.
• The cut pieces of wire is connected in parallel
• Total Resistance = $\bold{r_1}$

To find :

• Relation $\bold{\frac{R}{r_1}}$

Solution :

Resistance of each part = $\bold{\dfrac{R}{5}}$

Net resistance in parallel,$\bold{\frac{R}{r_1}}$

$\rightarrow$$\bold{\frac{1}{r_1}}$ = $\bold{\frac{1}{R/5}}$ + $\bold{\frac{1}{R/5}}$ + $\bold{\frac{1}{R/5}}$ + $\bold{\frac{1}{R/5}}$ + $\bold{\frac{1}{R/5}}$

$\rightarrow$$\bold{\dfrac{1}{r_1}}$ = $\bold{\dfrac{5}{R}}$ + $\bold{\dfrac{5}{R}}$ + $\bold{\dfrac{5}{R}}$ + $\bold{\dfrac{5}{R}}$ + $\bold{\dfrac{5}{R}}$

$\rightarrow$$\bold{\dfrac{1}{r_1}}$ = $\bold{\dfrac{5+5+5+5+5}{R}}$

$\rightarrow$$\bold{\dfrac{1}{r_1}}$ = $\bold{\dfrac{25}{R}}$

$\rightarrow$$\bold{\dfrac{r_1}{1}}$ = $\bold{\dfrac{R}{25}}$

•°• $\bold{\dfrac{R}{r_1}}$ = 25