Class-10 Goyal Brothers Electric Circuit, Resistance and Ohm’s Law ICSE Physics Solutions Ch-8

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Class-10 Goyal Brothers Electric Circuit, Resistance and Ohm’s Law ICSE Physics Solutions Ch-8 . We Provide Step by Step Answer of Exercise, MCQs, Numerical Problem Questions of Exercise-1 Goyal Brothers Electric Circuit, Resistance and Ohm’s Law ICSE Class-10 . Visit official Website CISCE  for detail information about ICSE Board Class-10 Physics.

Electric Circuit, Resistance and Ohm’s Law Class-10 Goyal Brothers ICSE Physics Solutions Ch-8


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Exercise-1

 MCQ -1

Numerical Problem-1

Exercise-2

 MCQ -2

Numerical Problem-1


Exe-1 Class-10

Electric Circuit, Resistance and Ohm’s Law Goyal Brothers ICSE Physics Solutions Ch-8

Page-131

Question 1.
In which direction conventional current and electronic current flow from a source of electricity ?

Answer:
Electronic current is always in opposite direction to conventional current.

  1. When both the bodies are positively charged and are in contact the body having more +ve charge is at higher potential conventional current from A to B + 100 > + 70 from higher potential to lover potential.
    ∴ Electronic current from B to A.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 12.2
  2. When both A and b negative charge conventional current: from higher potential (—70) to lower potential (—100) i.e. from B to A electronic current: from A to B.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 12
  3. When A has +ve charge of 100 units and B has —ve charge of 70 units Conventional Current: from A to B Higher potential to lover potential a electronic current from b to A.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 12.1

 

 

 

 

 

Question 2.
Define electric potential. State its practical unit and define it.
Answer:
Electric Potential : “Is the electrical state of a conductor which determines the direction of flow of charrge when two conductors are either in contact or joined by a metallic wire.”
Or
ELECTRIC POTENTIAL “at a point is the work done in moving unit positive charge from infinity to that point in an electric field.
S.I. unit is volt.
VOLT : “If work done in moving 1 coulomb of charge from one point to other is 1 joule, the potential-difference between two points is said to be 1 volt.”

Question 3.
Define quantity of charge. States its practical unit and define it
Answer:
Quality of charge is “The number of charges (electrons) which drift from a higher to a lower potential is called quantity of charge.”
Particle unit of charge is coulomb.
Coulomb: “Flow of 6.25 × 1018 electrons through a conductor constitute 1 coulomb.”
Or
“Charge carried by  6.25 × 1018 electrons is called 1 coulomb.”

Question 4.
Define electric current State its practical unit and define it
Answer:
Electric current: “Rate of flow of charge”. I = Q/t = ne/t
S.I. Unit → Ampere (A)
Ampere : “When a charge of 1 coulomb passes in 1 second current flowing is one ampere.”

Question 5.
State two multiples and two submultiples of the unit of electric potential and electric current
Answer:
Multiple units of :
Electric potential is

  1. ( Kv ) kilovolt = 103 V
  2. Megavolt (Mv) = 10V

Electric current :

  1. KA (kilo-ampere) = 1000 A
  2. MA (Mega-ampere) = 106

Sub multiple units of:
Electric potential:

  1. mV millivolt = 10-3 V
  2. µv = microvolt = lO-6 V

Electric current :

  1. mA = milliampere = 10-3 A
  2. µA = micro-ampere = 10-6 A

Question 6.
What do you understand by the terms potential difference? State its practical unit
Answer:
Potential difference: “Is the amount of work done in moving a unit positive charge from one point to other.” Practical unit – volt.

Question 7.
Define
(a) open electric circuit
(b) closed electric circuit.
Answer:
(a) Open electric circuit : “An electric circuit in which low of current stops, because of an open switch is called an open_electric circuit.
(b) Closed electric circuit : “An electric circuit in which a current flows continuously, because the switch is closed is called a closed electric .
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 14

Question 8.
What do you understand by the term electric resistance? state its practical unit.
Answer:
“The obstruction offered to the flow of current by a conductor (wire) is called its RESISTANCE.”
S.l. UNIT — is 1/OHM or (Ohm)-1

Question 9.
What do you understand by the term electric conductance? State its practical unil
Answer:
“The reciprocal of resistance is called electric CONDUCTANCE”. i.e.
Conductance = 1/Resistance
S.l. Unit of Conductance Ω-1 or OHM-1

Question 10.
What is a superconductor ? Name two materials and the temperature at which they become superconductors.
Answer:
SUPER CONDUCTOR : “The substances which lose resistance when they are cooled to very low temperature (nearly absolute zero) are called super conductors, e.g. mercury at 4.12 k, LEAD. TIN, VANADIUM etc. and this phenomenon is called SUPER CONDUCTIVITY. The temperature at which they become super-conductors is called CRITICAL TEMPERATURE.

Question 11.
State the laws of resistance.
Answer:
Laws of resistance :

  1. Resistance of a conductor is directly proportional to its length R α l
  2. Resistance of a conductor is Inversely proportional to its area of cross-section R α 1/a
  3. Resistance of a conductor depend upon its nature i.e. copper has less resistance than iron.
  4. Resistance of a conductor increases with increase in temperature i.e. resistance of filament of a bulb is more when lighted as compared to when it is not lighted.

Question 12.
Define specific resistance and state its unit in CGS and SI system.
Answer:
Specific resistance: “Is the resistance of a wire of that material of unit length and unit area of cross-section.”
Unit:
In C.G.S. system → [Ω – cm] ohm – cm
In S.I. system → [Ω – m] ohm – metre.

Question 13.
Name two materials in each case whose resistance
(a) increases,
(b) remains the same and
(c) decreases with the rise in temperature.
Answer:
Two materials:

  1. Cu, iron, sp. resistance increases Tungston with increase in temp.
  2. Metallic alloys like eureka, Manganin and German silver The sp. resistance remains the same with rise in temp.
  3. Carbon and Rubber. Resistance decrease with increase in temperature.

Question 14.
Give two differences between the electric resistance and electric specific resistance of a material
Answer:
Two differences between resistance and sp. resistance.
Resistance :

  1. S.I. unit is ohm (Ω)
  2. It is measured as ratio of pot. difference at the ends of a conductor to the current flowing through the conductor.

Resistivity or sp. Resistance :
S.I. unit [Ω-m] ohm. metre. It is measured as the resistance offered by a conductor of unit length and unit area of cross-section.


 Multiple Choice Questions-1

Electric Circuit, Resistance and Ohm’s Law Class-10 Goyal Brothers  ICSE Physics Solutions

Question 1.
The graph between V/I for a conductor is a straight line. The slope of the graph represents :
(a) resistivity
(b) resistance
(c) electric potential
(d) none of these
Answer:
(b) resistance

Question 2.
Two conductors A and B have 500 and 100 units of . negative charge when the conductors are connected by
an electric wire the conventional current flows from :
(a) A to B
(b) B to A
(c) Current does not flow
(d) none of these
Answer:
(b) B to A

Question 3.
A conductor at 4.2 K is found to offer no resistance. Such a conductor is called
(a) zero conductor
(b) superconductor
(c) absolute conductor
(d) none of these
Answer:
(b) superconductor

Question 4.
Which of the following is non-ohmic resistance ?
(a) Copper wire
(b) Brass wire
(c) Copper wire wound on an electromagnet
(d) Constantan wire
Answer:
(b) Brass wire

Question 5.
Which of the following an ohmic resistance ?
(a) Diode valve
(b) Filament of a bulb
(c) Carbon are light
(d) Manganin wire
Answer:
(d) Manganin wire

Question 6.
A conductor has a resistivity of 2.63 × 10-8 Ω m at 20° C. If the temperature of conductor is raised to 200°C, its resistivity will :
(a) increase
(b) decrease
(c) remain unaffected
(d) none of these
Answer:
(a) increase

Question 7.
Amongst the following substance, the resistance will decrease with the increase in temperature in case of:
(a) copper
(b) carbon
(c) brass
(d) nichrome
Answer:
(b) carbon


Numerical Problem on Specific Resistance

Electric Circuit, Resistance and Ohm’s Law Class-10 Goyal Brothers  ICSE Physics Solutions

Page-132

Practice Problems : 1

Question 1.
A wire of resistance 4.5 Ω and length 150 cm, has an area of cross-section of 0.04 cm-2. Calculate sp. resistance of the wire.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 18

Question 2.
A wire of length 40 cm and area of cross-section 0.1 mm2 has a resistance of 0.8 fl Calculate sp. resistance of the wire.
Answer:
I = 40 cm, area of cross-section a = 0.1 m m2 = 0.1/100 c m2 R = 0.8 Ω
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 19

Practice Problems : 2

Question 1.
Resistance of a conductor of length 75 cm is 3.25 Ω. Calculate the length of a similar conductor, whose resistance is 13.25Ω.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 19.1

 

Question 2.
A conductor of length 85 cm has a resistance of 3.750. Calculate the resistance of a similar conductor of length 540 cm.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 19.2

 

 

 

 

Practice Problems : 3

Question 1.
A resistance wire made from German silver has a resistance of 4.250. Calculate the resistance of another wire, made from same material, such that its length increases by 4 times and area of cross-section decreases by three times.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 19.3

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 20

Question 2.
A nichrome wire of length l and area of cross-section a/ 4 has a resistance R. Another nichrome wire of length 31 and area of cross-section a/2 has a resistance of R1 Find the ratio of R, : R.
Answer:
As both the wire are made of same material, have same sp. resistance
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 20.1

 

 

 


Exe-2 Class-10

Electric Circuit, Resistance and Ohm’s Law Goyal Brothers ICSE Physics Solutions Ch-8

Page-138,139

Question 1.
(a) Define series circuit.
Answer:
(a) Series circuit: “Resistances are said to be connected in series if same current flows through them and the resistance are connected end to end.
(b) State three characteristics of a series circuit Ans. Characteristics of series circuit:
Answer:

  1. Same current flows through each resistance.
  2. V = V1 + V2 + V3 …. i.e. total potential drop is the sum of individual resistances.
  3. When we want higher resistance, connect them in series. (Resistance is more than individual resistances).

Question 2.
(a) Define parallel circuit.
Answer:
Parallel circuit : “Resistances are said to be connected in parallel if one end of each is connected at a common terminal and other end of each at other common terminal and they have a common pot. difference.”
(b) State three characteristics of a parallel circuit.
Answer:
Characteristics of parallel circuit:

  1. Pot. difference of each resistance is same.
  2. Current divides [I = I1 + I2 + I3…]
  3. I/R = I/R1 + I/R2 + I/R3 reciprocal of total resistance is the sum of reciprocals of individual resistances.
  4. Total resistance is less than the least of individual resistances.

Question 3.
(a) Stale Ohm’s law.
Answer:
Ohm’s Law : “Physical conditions like temp. remaining the same potential across the ends of a conductor is directly proportional to the current flowing”.

(b) What are the limitations of 0hm‘s law?
Answer:
Ohms’Law is obeyed only when temperature remains constant.

Question 4.
How will you verify Ohm’s law by voltmeter, ammeter method?
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 21

Verification of Ohm’s Law: Use the circuit as shown taking case the +ve of voltmeter and +ve of Ammeter should be connected to the +ve of battery and voltmeter in parallel key is closed and Rheostat is set to get the minimum reading in Ammeter and voltmeter. The rheostat is then gradually moved
and each time value of A and V are noted. The ratio of V/I is always found constant. This verified ohm’s law.

Question 5.
How will you verify Ohm’s law by potentiometer method?
Answer:
Potentiometer method to verify
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 22
Connect the potentiometer as shown, close the key and record the potential difference by pressing the jockey at 10 cm intervals of length of the potentiometer wire. Repeat the experiment for six different lengths of potentiometer wire and record the corresponding pot. differences.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 22.1
Find the ratio between potential difference and length. It is found.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 22.2
V/l = constant (I) (where I is current and the current in series circuit is constant quantity)
But l α R {By law of resistance}
V/R = I or I = V/R
Hence ohm’s law is verified.

Question 6.
What are ohmic resistances ? Given two examples.
Answer:
Ohmic resistances : “Conductors which obey ohm’s law are called ohmic resistances.”
Two examples : Vanadium, all pure metals like Cu, Al, etc.

Question 7.
What are non-ohmic resistances ? Give two examples.
Answer:
Non-ohmic resistances: “The resistances which do not obey ohm’s law are called non-ohmic resistances.”
Two examples : Diode valve, triode valve, transistors, filament of a bulb.

Question 8.
Derive an expression for three resistances connected in series.
Answer:
EQUIVALENT RESISTANCE OF RESISTORS
(i) in PARALLEL :
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 23

Let three resistors are connect in such away that one end of each R1, R2, R3 is connected at a common terminal (X) and the other end of each at common terminal (Y) through a battery. So that potential difference of each resistor is V and current I at X divides itself and I1, I2 and I3 flows through R,, R, and R3 respectively and again combine at y and current I flows further.
I = I1 + I2 + I3 …(i) we know that
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 24
EQUIVALENT or RESULTANT resistance of parallel conductors
The reciprocal of EQUIVALENT RESISTANCE is equal to the sum of the reciprocals of individual resistors.
EQUIVALENT RESISTANCE OF RESISTORS In (ii)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 24.1
Let three resistors R1, R2, and R3, be connected in series i.e. resistors are joined one after the other as shown and same current I passes through each and each resistors has potential difference say V,, V, and V3 so that total p.d. between A and D terminals is V
V = V1 + V2 + V3 …(i) V = IR
IR = I R1 + IR2 + IR3 V1 = IR1
IR = I[R1 + R2 + R3] V2 = IR2
V3 = IR3 put in (i)
R = R1 + R2 + R3
i. e. EQUIVALENT RESISTANCE of resistors in series is the sum of their individual resistance.

Question 9.
Derive an expression for three resistances connected in parallel.
Answer:
To derive relation for resultant resistance of three resistances in parallel, consider three resistances
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 25

  1. r1, r2, r3 One end of each is connected to common terminal X and other end of each at common terminaly.
  2. Here current I divides into I1, I2, I3, flowing through r1, r2, r3 and potential difference is v for all resistances.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 25.1
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 26

Question 10.
What do you understand by the term internal resistance of a cell ?
Answer:
INTERNAL RESISTANCE OF A CELL :
“The resistance offered by the electrolyte inside the cell to the flow of current is called the internal resistance of cell.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 26.1

Question 11.
State the factors on which internal resistance of a cell depends.
Answer:
Factors effecting the internal resistance of a cell :

  1. Surface area of electrodes larger surface area, lesser is the internal resistance.
  2. Distance between electrodes : more the distance, more is the internal resistance.
  3. Temp, of electrolyte r α 1/T
  4. Higher the concentration of electrolyte greater is internal resistance.

Question 12.
What is the difference between emf and terminal voltage of a cell ?
Answer:
Difference between e.m.f. and terminal voltage :
Terminal voltage : When current is drawn from a cell i.e. the cell is in a closed circuit, the potential differences between the electrodes (terminals) of a cell is called terminal voltage,
e.m.f. : “when no current is drawn from a cell i.e. when the cell is in open circuit, the pot. difference between the terminals of the cell is called electromotive force, (e.m.f.).


 Multiple Choice Questions-2

Electric Circuit, Resistance and Ohm’s Law Class-10 Goyal Brothers  ICSE Physics Solutions

Question 1.
In a series circuit:
(a) p.d. across all resistors is same
(b) current flowing through all resistors is same
(c) The combined resistance of all resistors is less than individual resistors.
(d) none of the above
Answer:
(b) current flowing through all resistors is same

Question 2.
In a parallel circuit:
(a) p.d. across all resistors is same
(b) current flowing through all resistors is same
(c) the equivalent resistance of all resistors is more than any of the individual resistors
(d) none of the above
Answer:
(a) p.d. across all resistors is same

Question 3.
Two resistors of 2 Ω. each are connected in a parallel. The equivalent resistance is :
(a) less than 2 Ω but more than 1Ω
(b) one ohm
(c) four ohm
(d) between 4 Ω and 2 Ω
Answer:
(b) one ohm

Question 4.
A new cell is marked 1.5 V. When connected to an external resistance, the voltmeter connected to its terminals reads 1.2 V. The drop in potential across the terminals of the cell is due to the :
(a) internal resistance of cell
(b) external resistance .
(c) both (a) and (b)
(d)none of these
Answer:
(a) internal resistance of cell

Question 5.
A potentiometer is connected to a cell through switch in series. To one end of the potentiometer is attached a voltmeter with the help of connecting wire and a jockey. When the jockey is moved over the potentiometer wire from zero end to 100 cm the reading shown by voltmeter is likely to :
(a) decrease
(b) increase
(c) does not change
(d) none of these
Answer:
(b) increase

Question 6.
When the current is drawn from a cell in a closed circuit, the potential difference between the terminals of cell is called :
(a) e.m.f.
(b) p.d.
(c) terminal voltage
(d) both (a) and (b)
Answer:
(c) terminal voltage



Numerical Problem on Resistance

Electric Circuit, Resistance and Ohm’s Law Class-10 Goyal Brothers  ICSE Physics Solutions

Page-139,140,141

Practice Problems : 1

Question 1.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 28

Calculate the equivalent resistance

  1. between points A and B
  2. between points A and C.

Answer:
Resistance 6Ω, 3Ω and 2Ω are in parallel between A and B

  1. Equivalent resistance between AB is R1  ∴ R1 = 1Ω Now combination R1 and IΩ of BC are in series
  2. Now between points A and C R = R1 + 1 = 1 + 1 = 2Ω

Question 2.
In figure, calculate equivalent resistance between points

  1. A and B
  2. B and C
  3. A and C.

Answer:
Equivalent resistance between AB 4Ω and 12 Ω are in parallel
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 29

2.Between B and C
Equivalent Resistance R2 between BC
3.Between A and CD
Now R1 and Rand in series
Equivalent Resistance between A and C
1 R = R1 + R2 = 3 + 4 = 7Ω

Practice Problems : 2

Question 1.
Calculate the equivalent resistance between points

  1. B and E (ii) A and F.
  2. A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 30
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 30.1
    1.Equivalent Resistance between BE, (1Ω, 2Ω, 3Ω are in series) isR1

R1 = 1 + 2 + 3 = 6Ω
R1 is in parallel to R2 = 3Ω
their resultant R.3
1/r3 = 1/r1 +1/r2 = 1/6 + 1/3 = 1+2/6 = 1/6 Between BE
:.R= 2Ω
(ii) Equivalent resistance between A and F
i.e. R4, R2, Rs are in series
3Ω, 2Ω and 3Ω are in series
R= 3 + 2 + 3 = 812
Answer:

Question 2.
Calculate the equivalent resistance of circuit diagram shown in Fig. below.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 31
Resistances A and B are in series
R= 4 + 8 = 12Ω
Resistance C and D are in series
R2 = 1.5 + 4.5 = 6Ω
Now R1, E and R2 are in parallel
:. Equivalent resistance between F and G
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 31.1

Practice Problems : 3

Question 1.
Equivalent resistance of circuit diagram is 6Ω. Calculate the value of x.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 32

Question 2.
Equivalent resistance of circuit diagram is 5Ω. Calculate the value of x.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 32.1
Answer:
Since Equivalent resistance of 4Ω and parallel combination is 5Ω and 4Ω and parallel combination are in series.
Resistance of parallel combination is = 5 – 4= lΩ
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 33


Numerical Problems on Ohm’s Law

Electric Circuit, Resistance and Ohm’s Law Class-10 Goyal Brothers  ICSE Physics Solutions

Page-141,142,143,144,145,146,147

Practice Problems : 1

Question 1.
A current of 0.2 A flows through a conductor of resistance 4.50. Calculate p.d. at the ends of conductor.
Answer:
Here I = 0.2 A, R = 4.5Ω
p.d. at the ends of conductor V = IR
V = 0.2 × 4.5 = 0.9 V

Question 2.
A bulb of resistance 4000 is connected to 200 V mains. Calculate the magnitude of current.
Answer:
R = 400Ω, V = 200 V
I = V/R 200/400 = 0.5 A

Question 3.
An electric heater draws a current of 5 A, when connected to 220 V mains. Calculate the resistance of its filament.
Answer:
I = 5 A, V = 220 V, R = ?
R = v/I = 220/5 = 44Ω

Practice Problems : 2

Question 1.
Four resistors of resistance 0.5 Ω, 1.5Ω, 4Ω and 6Ω are connected in series to a battery of e.m.f. 6 V and negligible internal resistance. Calculate :

  1. current drawn from the cell
  2. p.d. at the ends of each resistor.

Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 34

 

 

 

 

 

 

 

 

 

 

 

Question 2.            Class-10 Goyal Brothers Electric
Figure shows a circuit diagram having a battery of 24 V and negligible internal resistance. Calculate :

  1. reading of the ammeter,
  2. reading of V1, V2 and V3.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 34.1

Answer:
As 6Ω and 3Ω are in parallel
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 35

Practice Problems : 3

Question 1.
Three resistors of 6Ω, 2Ω and x are connected in series to a cell of e.m.f 3/2 V, when the current registered in circuit is 1/6 A. Draw the circuit diagram and calculate value of x
Answer:
Circuit diagram:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 35.1

Question 2.
Carefully study the circuit diagram in figure and calculate the value of resistor x.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 36
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 36.1

 

 

 


A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 37

 

 

 

 

Practice Problems : 4

Question 1.              Class-10 Goyal Brothers Electric
Three resistors of 4Ω, 6Ω. and 12Ω are connected in parallel The combination of these resistors is connected in series to a resistance of 2Ω and then to a battery of e.m.f 6 V and negligible internal resistance.
(a) Draw the circuit diagram
(b) Calculate the current in main circuit
(c) Calculate the current in each of the resistors in parallel
Answer:
(a) Circuit diagram
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 37.1

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 38

Question 2.
Study the circuit diagram in figure carefully and calculate:
(a) current in main circuit
(b) current in each of the resistors in parallel circuit.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 38.1
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 38.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 39

Practice Problems : 5

Question 1.
Figure shows a circuit diagram containing 12 cells, each of e.m.f 1.5 V and intenal resistance 0.25Ω Calculate:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 39.1
(a) Total internal resistance
(b) Total e.m.f.
(c) Total external resistance
(d) Reading shown by the ammeter
(e) Current in 12Ω and 8Ω resistors
(f) p.d. across 2.2 resistor
(g) Drop in potential across the terminals of the cell
Answer:
Number cells in series = n – 12
(a) Total internal resistance of 12 cells = n r = 12 × 0.25 = 3Ω
(b) Total e.m.f. = 12> <1.5 = 18 v
(c) Total external resistance : 4 + 8 = 12 Ω in series
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 39.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 40
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 40.1

Question 2.
Four cells, each of e.m.f 2 V and internal resistance 0.2 Ω each are connected in series to form a battery. This battery is connected to an ammeter, a resistance 1.2 and then to a set of resistance of 4 Ω, 6 Ω and 12 Ω in parallel to complete the overall circuit in series.
(a) Draw circuit diagram of arangment.
(b) Calculate total internal resistance
(c) Total e.m.f.
(d) Current recorded by ammeter.
(e) Current flowing through 6 wire in parallel.
(f) Drop in potential across the terminals of the battery.
Answer:
(a) Circuit diagram :
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 41
(b) Total internal resistance r = 4 × 0.2 = 0.8 Ω
(c) Total e.m.f. = 4 × 2 = 8 v
(d) Current recorded by ammeter ?
R1 across CD 1/R1 = 1/4 + 1/6 + 1/12 = R1 = 2 Ω
Total resistance (0.8 + 1.2 + 2) = 4 Ω
I = v/R = 8/4 = 2a
(e) Vacross CD = I R1= 2 × 2 = 4 V
Current through 6 Ω = v/R = 4/6 = 0.67 A
(f) droop in potential across the terminals of the battery E – V = Ir = 2 × 0.8 = 1.6 V

Practice Problems : 6

Question 1.    Class-10 Goyal Brothers Electric
Two cells, each of e.m.f 1.5 V and internal resistance 1 Ω are connected in parallel, to form a battery. The battery is connected to an externari resistance of 0.5 Ω and two resistances of 3 Ω and 1.5 Ω in parallel.
(a) Draw the circuit diagram.
(b) Calculate the current in main circuit.
(c) Calculate the current in 1.5 Ω resistor.
(d) Calculate the drop in potential across the terminal of the battery.
Answer:
(a) Circuit diagram
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 42
(b) Current in the main circuit
Total internal resistance of two parallel cells I/R1 = 1 + 1
R = 1/2 = O.5 Ω
Effective resistance between PQ = 1/R2 = 1/3 + 2/3
R2 = 1 Ω
Total resistance of circuit R = R1 + R2+ external resistance
R= O.5 + I + O.5 = 2Ω, V= 1.5 V
current in main circuit
I = V/R = 1.5/2 = O.75 A
(c) Current in 1 .5 Ω resistor
p.d. between PQ = IR2
V1 = 0.75 × 1
V1 = 0.75 V
V1 = 0.75
I1= V1/1.5 = 0.75/1.5 = O.5 A
(d) Drop in potential across the terminals of battery E – V = Ir = 0.75 × 0.5 = 0.375 V

Question 2.
Four cells, each of e.m.f. 1.5 V and internal resistance 2 Ω. each are connected in parallel to form a battery. The battery is connected to an external resistance of 0.5 Ω. and three resistances of 12 Ω, 6 Ω and 4 Ω. in parallel.

  1. Draw the circuit diagram.
  2. Calculate current in main circuit
  3. Calculate current in 4 D resistor.
  4. Calculate drop in potential across the terminals of  battery.

Answer:

1.circuit diagram is drawn
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 43
4. Drop in potential across the terminals of battery

E – V = 1r
= 0.5 × 0.5 = 0.25 V

Practice Problems : 7

Question 1.
A cell of e.m.f 1.5 V, records a p.d. of 1.35 V, when connected lo an external resistance R, such that current flowing through circuit is 0.75 A. Calculate the value of R and internal resistance of cell
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 44

Question 2.
In figure a current of 1 A flows through the circuit, when p.d. recorded at the ends of parallel resistors is 1 volt. Calculate the value of R and r.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 44.1
Answer: For parallel resistors
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 45
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 45.1

Practice Problems : 8

Question 1.
A cell of e.m.f 1.8 V is connected to an external resistance of 2 Ω, when p.d. recorded at the ends of resistance is 1.6 V. Calculate the internal resistance of the cell.
Answer:
I = V/R
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 45.2

Question 2.
Study ttitel circuit diagram in Fig. 8.44, and hence, calculate the internal resistance of cel                                      A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 46

Practice Problems 9

Question 1.  Class-10 Goyal Brothers Electric
A cell, when connected to an external resistance of 4.5 Ω shows a p.d of 1.35 V. If 4.5 Ω resistance is replaced by 2.5Ω resistance the p.d drops to 1.25 V. Calculate:
(a) em.f.,
(b) internal resistance of the cell
Answer:
Let ‘E’ be the e.m.f and ‘r’ the internal resistance
Case(l) r = R [e-v]/v
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 47

Question 2.
Study the figures carefully and hence calculate the value of E and r.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 47.1
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electric Circuits, Resistance & Ohm’s Law 48

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