Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Class 10 Physics Solutions By Lakhmir Singh & Manjit Kaur

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Class 10 : Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

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Lakhmir Singh Physics Class 10 Solutions Page No:38

Question 8:
Show how you would connect two 4 ohm resistors to produce a combined resistance of
 (a) 2 ohms
 (b) 8 ohms.

Solution :
(a) By connecting in parallel: Since equivalent resistance will be
1/ R = 1/4 + 1/4 = 2/4 = 1/2
Therefore, R = 2 ohm
(b) By connecting in series : Since equivalent resistance will be
R = 4 ohm + 4 ohm = 8 ohm.

Question 9:
Which of the following resistor arrangement, A or B, has the lower combined resistance ?

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev
Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Solution :
Resistance of arrangement A is 10 ohm.
Combined resistance of arrangement B is caculated as follows:
1/R = 1/10 + 1/1000 = (100+1)/1000
R = 1000/101 = 9.9 ohm
Therefore, arrangement B has lower combined resistance.

Question 10:
A wire that has resistance R is cut into two equal pieces. The two parts are joined in parallel. What is the resistance of the combination ?
Solution :
Resistance of each part is R/2.
Resultant resistance R’ is given by
1/R’ = 2/R+2/R
R’=R/4.

Question 11:
Calculate the combined resistance in each case :

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev
Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev
Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Solution :

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev
Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev
Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Question 12:
Find the current in each resistor in the circuit shown below :

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Solution :

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Question 13:
Explain with diagrams what is meant by the “series combination” and “parallel combination” of resistances. In which case the resultant resistance is : (i) less, and (ii) more, than either of the individual resistances ?
Solution :

(i) Series combination 
When two or more resistances are connected end to end consecutively, they are said to be connected In series combination. The combined resistance of any number of resistances connected in series in equal to the sum of the individual resistances. 

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

The resultant resistance is more than either of the Individual resistances. 

(ii) Parallel combination 
When two or more resistances are connected between the same two points, they are said to be connected in parallel combination. The reciprocal of the combined resistance of a number of resistances connected In parallel Is equal to the sum of the reciprocals of all the Individual resistances.

Question 14:
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor ?

Solution :

R1 = 0.2ohm, R2 = 0.4ohm, R3 = 0.3ohm,R4 = 0.5ohm. R5 =12ohm, V = 9V
Resultant resistance = R1+ R2+ R3+ R4+ R5
R = 0.2+0.4+0.3+0.5+12 =13.4ohm
Thus the current flow through 12ohm resistance will be.V/R
I = 9/13.4
I = 0.67amp. 

Question 15:
An electric bulb of resistance 20 Ω and a resistance wire of 4 Ω are connected in series with a 6 V battery. Draw the circuit diagram and calculate :
 (a) total resistance of the circuit.
 (b) current through the circuit.
 (c) potential difference across the electric bulb.
 (d) potential difference across the resistance wire. 

Solution :

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

(a) Total resistance of the circuit=R1-FR2=20+4=24ohm

(b) We know that

V = I R

Therefore.

6 = 1 x 24

I = 6/24 = 0.25amp

(c) p.d. across bulb = I R1= 0.25X20 = 5V

(d) p.d. across resistance wire= I R2 = 0.25X4 = 1V 


Question 16:
Three resistors are connected as shown in the diagram.

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Through the resistor 5 ohm, a current of 1 ampere is flowing,
 (i) What is the current through the other two resistors ?
 (ii) What is the p.d. across AB and across AC ?
 (iii) What is the total resistance ?

Solution :

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

According to the diagram.

(i) Total current I = 1 amp is entering the parallel combination of R1 and R2. Let I1 current flow through R1 and I2 current flow through R2. Then 

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

(ii) p.d. across AB - IR3 = 1 x 5 - 5V

Equivalent resisyance between B and C is

1/R' = 1/R1 + 1/R2 = 1/10 + 1/15
1/R' = 5/30
R' - 6 ohm

Total resistance between A and C is R = 5-6 = 11 ohm p.d. across AC = IR = 1x11= 11V 
(iii) Total resistance = R3 + R' = 5+ 6 =11 ohm 

Question 17:
For the circuit shown in the diagram below :

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

What is the value of :
 (i) current through 6 Ωresistor ?
 (ii) potential difference across 12 Ω resistor ?

Solution :

As per the circuit V = 4 V

Total resistance in line 1 = R1 = 6 + 3 = 9 ohm
Total resistance in line 2 = R2 = 12 + 3 = 15 ohm 

(i) Current through 60 resistor = current through line   Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

p.d. across line 2 is 4V 

Solutions of Electricity (Page No- 38) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev         

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