Question

Derive the effective resistance of three resistors R 1 , R 2 & R 3 when connected in series, with the help of a labelled diagram.

Answer 1

Answer:

r = r1+r2+r3

Explanation:

ncert given answer in book

Answer 2

Answer:

10.33 Ω

Explanation:

Derivation for effective resistance is series.

In series combination :

Current i is same in all resistance

Potential difference isn't same!

= > V = V₁ + V₂ + V₃ ... ( i )

Let we have three resistors R₁ , R₂ and R₃

From ohm's law we know :

V = I R

= > Using for ( i ) we get :

I R_e = I R₁ + I R₂ + I R₃

= > R_e = R₁ + R₂ + R₃

Now finding effective resistance given in question :

Here 5 Ω and 10 Ω are in parallel :

R_net = R₁ R₂ / R₁ + R₂

R_e = 7 + ( 50 / 15 )

= > R_e = ( 105 + 50 ) / 15

= > R_e = 155 / 15  Ω = 10.33 Ω

Hence we get required answer!