Solutions of Electricity (Page No- 43) - Physics By Lakhmir Singh, Class 10 | Extra Documents, Videos & Tests for Class 10 PDF Download

Lakhmir Singh Physics Class 10 Solutions Page No:43

Question 43:
Show with the help of diagrams, how you would connect three resistors each of resistance 6 Ω, so that the combination has resistance of (i) 9 Ω (ii) 4 Ω.
Solution :

Resultant resistance for parallel circuit=R

1/R = 1/6 + 1/6

1/R = 2/6 R = 3

Effective resistance = 6 + 3 = 9ohms  

Resultant resistance for each parallel circuit= R

1/R = 1/6+1/6+1/6

1/R = 3/6

R = 2

Therefore effective resistance = 2+2 = 4ohms. 

Question 44:
Two resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.
Solution :
Two resistances when connected in series, resultant value is 9ohms.

Two resistances when connected in parallel, resultant value is 2ohms.

Let the two resistances be R₁ and R₂.

If connected in series, then

9 = R₁₊ R₂

R₁ = 9-R₂

If connected in parallel, then

1/2 = 1/R₁+1/R₂

From above equations we get that

1/2=(R₁₊R₂)/R₁R₂

1/2 = 9/(9-R₂) R₂

9R₂-R₂² = 18

R₂²-9R₂+18 = 0

(R₂-6) (R₂-3)=0

R₂ = 6,3

So if R₂ = 6ohms, then R₁ = 9-6 =3ohms.

If R₂ = 3ohms, then R₁= 9-3 = 6ohms. 

Question 45:
A resistor of 8 ohms is connected in parallel with another resistor X. The resultant resistance of the combination is 4.8 ohms. What is the value of the resistor X ?
Solution :

Given:

A resistor of 8ohm is connected in parallel with a resistor of X.

And resultant is 4.8.

Then  X = ?

We know that for parallel case

1/R = 1/R1+1/X
1/4.8 = 1/8+1/x
1/4.8 - 1/8 = 1/x

After solving we get that

X = 12ohms 

Question 46:
You are given three resistances of 1,2 and 3 ohms. Show by diagrams, how with the help of these resistances you can get:
(i) 6 Ω (ii) [latex s=2]\frac{6}{11}[/latex] Ω (iii) 1.5 Ω
Solution :

    Equivalent resistance = 10 + 20 + 30 = 60 
   Equivalent resistance 
 Equivalent resistance of first line = 1Ω + 2Ω = 3Ω
                                                                                        Resistance of the second line = 3 0 Equivalent resistance