Question

find the equivalent resistance of circuit if n equal resistances are connected in 1 ) series 2) p
arallel​

Answer 1

Answer:

Explanation:

Let the resistance of each resistor be R.

Series connection :

Equivalent resistance (maximum) R  

s

​  

=R+R+....+n terms

⟹ R  

s

​  

=nR

Parallel connection :

Equivalent resistance (minimum)  

R  

p

​  

 

1

​  

=  

R

1

​  

+  

R

1

​  

+...+n terms

We get  

R  

p

​  

 

1

​  

=  

R

n

​  

 

⟹ R  

p

​  

=  

n

R

​  

 

Thus ration of maximum to minimum resistance  

R  

p

​  

 

R  

s

​  

 

​  

=  

R/n

nR

​  

 

⟹  

R  

p

​  

 

R  

s

​  

 

​  

=n  

2

Answer 2

Explanation:

1) Let the number of resistances be n , the current through the circuit be I .

In series current remain same and voltage divide.

Therefore ,

V= V(1)+ V(2) + V(3) + ..... V(n)

From Ohm's Law :

V = IR

Putting this -

IR = I(R1) + I(R2) + I(R3) +..... I(R(n))

(I)R = (I) { (R1) + (R2) + (R3) ..... + (R(n))

* I cancel *

R = R1 + R2 + R3 + ..... R(n)

2) Let the number of resistances be n , the potential difference across the circuit be V.

In parallel potential difference remain same and current divide.

Therefore ,

I= I(1)+ I(2) + I(3) + ..... I(n)

From OHM'S law :

I = V/ R

Putting this -

$\frac{v}{r } = \frac{v}{r1} + \frac{v}{r2} + \frac{v}{r3} + ........ \frac{v}{r(n)} \\ (v)\frac{1}{r } = (v)(\frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} + ........ \frac{1}{r(n)}) \\ \frac{1}{r } = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} + ........ \frac{1}{r(n)}$