Question

Three conductors, each having Resistance of 3 ohms are joined in (i) series, (ii) in parallel, find the equivalent resistance in each case.

(Difficulty level - 2/5)

Answer 1

Explanation :-

Given :

Three conductors, each having resistance of 3 Ω are joined in (i) series, (ii) parallel.

To find :

Equivalent or effective resistance in each case.

Solution :

(i) As in series :

R_eff = R_1 + R_2 + R_3

Here, R_1 = R_2 = R_3 = 3 Ω

Hence, R_eff = (3 + 3 + 3)Ω

• 9 Ω [R_eff]

(ii) As in parallel :

1/R_eff = 1/R_1 + 1/R_2 + 1/R_3

Here, R_1 = R_2 = R_3 = 3 Ω

Hence, 1/R_eff = (1/3 + 1/3 + 1/3)Ω

• 3/3
• 1/1

1/R_eff = 1/1

• 1 Ω [R_eff)

Answer 2

As in series :

R_eff = R_1 + R_2+R_3

Here, R_1=R_2=R_3 = 30

Hence, R_eff = (3 + 3 + 3)

• 90 [R_eff]

(ii) As in parallel :

1/R_eff = 1/R_1+1/R_2 + 1/R_3 Hence, 1/R_eff = (1/3+1/3+1/3)Q

Here, R_1=R_2=R_3 = 30

3/3

1/1

1/R_eff = 1/1

• 10 [R_eff)