Solutions of Electricity (Page No- 42) - Physics By Lakhmir Singh, Class 10 | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 34:
(a) With the help of a circuit diagram, obtain the relation for the equivalent resistance of two resistances connected in parallel.
In the circuit diagram shown below, find :
(i) Total resistance.
(ii) Current shown by the ammeter A

Solution :

Suppose total current flowing the circuit is I. then the current passing through resistance R₁ wiII be I₁ and current passing through resistance R₂ will be l₂ .

Total current = I = I₁+I₂

Let resultant resistance of this parallel combination is R. By applying the Ohm's law to the whole circuit. we get that I = V/R

Since the potential difference across the both the resistances is same. so applying the Ohm's law to each resistance we get that

I₁ = V/R₁ 

I₂ = V/R₂

Putting these eq in the above one, we get that

V/R = V/R₁+V/R₂

1/R = 1/R₁+1/R₂

If two resistance are connected in parallel than. the resultant resistance will be

1/R = 1/R₁+1/R₂

(b) 

(i)Total reisitance = R

1/R=1/R₁+1/R₂

R₂ = 3+2=5ohms

R₁ = 5ohms

1/R=1/5+1/5

1/R=2/5

R=2.5ohms

Ili) Current flowing through the circuit

I = V/R = 4/(2.5) 

= 1.6amps 

Question 35:
(a) Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances R₁, R₂ and R₃ joined in parallel.
 (b) In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistors are identical.

With the switch K open, the ammeter reads 0.6 A. What will be the ammeter reading when the switch is closed ?
Solution :

Suppose total current flowing in the circuit is I. then the current passing through resistance R₁ will be I₁ Lcurrent passing through resistance R₂ will be I2 and current passing through resistance R₃ will bell I₃.

Total current = 1 = I₁+I₂+I₃

Let resultant resistance of this parallel combination is R. By applying the Ohm's law to the whole circuit. we get that

I=V/R

Since the potential difference across all the resistances is same. so applying the Ohm's law to each resistance we get that

l₁ = V/R₁ 
I₂ = V/R₂ 
I₃ = V/R₃

Putting these eqs. in the above one. we get

V/R = V/R₁V/R₂+ V/R₃ .

1/R = 1/R₁ + 1/R₂+1/R₃

If two resistance are connected in parallel, then the resultant resistance will be

1/R = 1/R₁+1/R₂+1/R₃

(b) If switch is open. then only upper two resistances (connected in parallel) are in the circuit

Effective resistance is 1/R_eq = 1/R+1/R = 2/R

R_eq = R/2

So the current.I = VA(R/2) =0.6A (given)

V/R :0.3A

When the switch closes, the third resistance also comes in the circuit. The effective resistance of the circuit becomes R/3

Hence. Current 1= V/(R/3) = 3 (V/R) = 3 x 0.3 .0.9 A