Question

three resistance 2 ohm , 4 ohm , and 6 ohm are connected in parallel calculate the equivalent resistance of this combination

Answer 1

Answer:

12/11 ohm is right answer

Answer 2

In parallel combination , the equivalent resistance of three resistors is given by :

$\tt{\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+ \dfrac{1}{R_3}}$

Or The reciprocal of equivalent resistance in parallel is the sum of the reciprocal of individual resistances .

Given :

• The three resistors are 2 Ω , 4 Ω and 6 Ω .

$\tt{\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} }$

Taking out the LCM of 2 , 4 and 6

$= \tt \dfrac{1 \times 6 + 1 \times 3 + 1 \times 2}{12}$

$\tt \: = \dfrac{6 + 3 + 2}{12}$

$\tt = \dfrac{11}{12}$

$\tt {R_p = \dfrac{12}{11}}$

Therefore , the equivalent resistance is 1.09 Ω.

⇢In parallel combination , equivalent resistance is smaller than the smallest resistor . Here the smallest resistor was of 2 Ω and the equivalent resistance is smaller than it .

⇢In series combination , the sum of all resistances gives the total resistance . So , the resistance is always larger than the largest resistance .