Question

Three resistors of 12 ,15 and 20 ohms are connected first in series and then in parallel. What is the equivalent resistance in each case - *

Answer 1

Solution :-

Given :-

$\bullet \ \sf R_1 = 12 \Omega \\\\\bullet \ R_2 = 15 \Omega \\\\\bullet R_3 = 20 \Omega$

When connected in series :-

Effective resistance,

$\bf R = R_1+R_2+R_3$

$\longrightarrow \sf R = 12+15+20 \\\\\longrightarrow R = 47 \Omega$

When connected in parallel :-

Effective resistance,

$\bf \dfrac{1}{R} = \dfrac{1}{R_1}+\dfrac{1}{R_2} +\dfrac{1}{R_3}$

$\longrightarrow \sf \dfrac{1}{R} = \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{20 } \\\\\longrightarrow \dfrac{1}{R} = \dfrac{1 \times 5 }{12 \times 5} +\dfrac{1 \times 4}{15 \times 4} + \dfrac{1 \times 3}{20 \times 3} \\\\\longrightarrow \dfrac{1}{R} = \dfrac{5}{60}+\dfrac{4}{60} +\dfrac{3}{60} \\\\\longrightarrow \dfrac{1}{R} = \dfrac{5+4+3}{60} \\\\\longrightarrow \dfrac{1}{R} = \dfrac{12}{60}\\\\\longrightarrow \dfrac{1}{R} = \dfrac{1}{5} \\\\ \longrightarrow R = 5 \Omega$

When connected in series, effective resistance = 47 Ω.

When connected in parallel, effective resistance = 5 Ω.