Important Questions
 

1. List out the apparatus required to verify that   is constant for a conductor. Explain the experiment with help of a diagram.
                                                               (OR)
'At constant temperature, the potential difference between the ends of a conductor is directly proportional to the current passing through it'. To verify this experimentally which apparatus are required? Explain it with the help of a diagram.    (AS 3) 4 Marks
A: AIM: To show that the ratio  is constant for a conductor.
Material required: 5 dry cells of 1.5 V each, conducting wires, an ammeter, a voltmeter, thin manganin spoke of length 10 cm and key.

Procedure:
‣ Connect a circuit as shown in the figure.
‣ Solder the conducting wires to the ends of the manganin spoke.
‣ Close the key. Note the readings of current (I) from ammeter and potential difference (V) from voltmeter in the table given below.

‣ Now connect two cells in the circuit and note the respective readings of ammeter and voltmeter in the above table.
‣ Repeat the above procedure using three cells and four cells and five cells respectively.
‣ Record the values of potential difference (V) and current (I) corresponding to each case in the above table.
‣ Find  for each set of values.
‣ We notice that   is a constant.
         V   I
‣ From this we can conclude that the potential difference between the ends of the manganin spoke is directly proportional to the current passing through it. The temperature of the manganin spoke is constant during the flow.

2. On what factors does the resistance of a conductor depend? How can you verify experimentally that any one of the factor affects the resistance. (AS 3) 4 Marks
A: The resistance of a conductor depends on nature of material and dimensions such as length, area of cross - section and temperature.
Verifying experimentally that resistance of a conductor depends on temperature:

‣ Take a bulb and measure the resistance of the bulb using a multimeter in open circuit. Note the value of resistance.
‣ Now connect the bulb in a circuit and switch on the circuit.
‣ After few minutes the bulb gets heated.
‣ Now measure the resistance of the bulb again with multimeter.
‣ The value of resistance of the bulb in second instance is more than the resistance of the bulb in open circuit.
‣ Here the increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.
‣ Thus the value of resistance of a conductor depends on the temperature.

3. How can you verify experimentally that the resistance of a conductor depends upon nature of material.   (AS 3) 4 Marks
A: Different metal rods are connected between P and Q

‣ Collect different metal rods of the same length and same cross sectional area like copper, aluminium, iron etc.
‣ Make circuit leaving gap between P and Q as shown in figure.
‣ Connect one of the metal rods between P and Q and switch on the circuit.
‣ Measure the current using an ammeter and note the value.
‣ Repeat this with other metal rods and measure electric current in each case.
‣ The values of current are different for different metal rods for a constant potential difference.
‣ From this we conclude that the resistance of a conductor depends on the material of the conductor.

4. Write the list of apparatus required to verify experimentally that the resistance of a conductor is inversely proportional to cross section Area. Write the procedure of the experiment with the help of a diagram.   (AS 3) 4 Marks
A: Required Materials: Battery, Ammeter, Key, Wires, Iron rods with different cross - section areas (lengths must be same).
Procedure:

‣ Collect iron rods of equal lengths but different cross - section areas.
‣ Make a circuit leaving gap between P and Q as shown in figure.
‣ Connect one of the rods between P and Q and measure the current using ammeter and note the values.
‣ Repeat this with the other rods and note the corresponding values of current in each case and note them.
‣ We will notice that the current flowing through the rod increases with increasing in the cross section area of the rod.
‣ Thus the resistance of the rod decreases with increasing the cross - section area.
‣ From this we can conclude that the resistance (R) of a conductor is inversely proportional to its cross-section area (A).
R     (at constant temperature and length of the conductor).

5. Show that effective resistance of a series combination in a circuit is equal to sum of their resistances. (AS 1) 4 Marks
A: Series connection: In a circuit, resistors connected end to end are said to be in series, if the same current exists in all of them through a single path.
                                       
* Consider three resistors R₁, R₂, R₃ are connected in series.
* V1, V2 and V3 are the potential difference across the resistances R₁, R₂ and R₃ respectively.
* Let I be the current flowing through each of the resistance.
* 'V' is the total potential difference in the circuit.
      .^.. V = V₁ + V₂ + V₃ .......... (1)
      By Ohm's law
      V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
      Substitute the values of V₁, V₂ and V₃ in equation (1)

V = IR₁ + IR₂ + IR₃ .......... (2)
Let R_eq is the equivalent resistance of the combination of resistors in series.
                
So, we have V = IR_eq
From equation ..... 2
         I R_eq = IR₁ + IR₂ + IR₃
         I(R_eq) = I(R₁ + R₂ + R₃)
            R_eq = R₁ + R₂ + R₃
* The sum of individual resistance is equal to their equivalent resistance when the resistors are connected in series.

6. Show that the reciprocal of the effective resistance of a parallel combination in a circuit is equal to the sum of the reciprocals of individual resistances.
                                                             (OR)
Explain the expression for the equivalent resistance of three resistors which are connected in parallel.     4 Marks
A: Parallel connection: In a circuit , resistors connected to common terminals are said to be in parallel, if identical potential difference exists across all of them.
                                    
* Consider three resistors R₁, R₂ and R₃ connected in parallel.
* Suppose a current I flows through the circuit when a cell of voltage 'V' is connected across the combination.

* The current I is divided as I₁, I₂, and I₃ which are flow through R₁, R₂ and R₃ respectively.
   I = I₁ + I₂ + I₃ ..... (1)
By Ohm's law 

Substituting the values of I₁, I₂ and I₃ in equation (1) 

Let R_eq be the equivalent resistance of the resistor is parallel.
^{V Then we get I =} 
From equation (2)

* The reciprocal of the equivalent resistance of the combination is equal to the sum of the reciprocals of the individual resistances.
 

7. State junction law and loop law and explain each law with one suitable example.    (AS 1) 4 Marks
A: i) Junction Law: At any junction point in a circuit where the current can divide, the sum of the currents in to the junction must equal. To the sum of the currents leaving the junction. This means that there is no accumulation of electric charges at any junction in a circuit.

I₁, I₄ and I₆ are the currents into the junction.
I₂, I₃ and I₅ are the currents leaving the junction according to junction law.
I₁ + I₄ + I₆ = I₂ + I₃ + I₅

ii) Loop Law: The algebraic sum of the increases and decreases in potential difference across various components of the circuit in a closed circuit loop must be zero.

Let us apply law to the circuit shown in figure.
i) For the loop ABCDEFA:
The resultant potential difference in the loop is
      -V₁ + I₁R₁ - I₂R₂ + V₂ = 0
ii) For the loop AFEDCBA:
The resultant potential difference in the loop is
       -V₂ + I₂R₂ - I₁R₁ + V₁ = 0

8. What do you understand about overload? Explain?
                                                                     (OR)
Why does the electrical appliances damages due to overload? How can we prevent the damages due to overload?     (AS 1) (AS 6) 4 Marks
A: Overload of Current:
       If the total current drawn from the mains is equal to the sum of the current passing through each device.
       If we add more devices to the household circuit the current drawn from the mains also increases. Then over heating occurs and may cause a fire. This is called overloading.
for Example:
* In our houses, the maximum current can be drawn from the mains is 5 - 20 A and potential difference is 240 V.
* When the current drawn from the mains is more than 20 A, caused overloading, devices may damages.
* From the given figure, if we switch on devices, such as heater, the current drawn from the mains exceeds the maximum limit 20 A.
* It causes fire.
* To prevent damages due to overloading, we connect an electric fuse to the household circuit, as shown in figure. In this arrangement, the entire current from the mains must pass through the fuse.
* The fuse consists of a thin wire of low melting point. When the current in the fuse exceeds 20 A, the wire will heat up and melt. The circuit then becomes open and prevents the flow of current into the household circuit. So, all the electric devices are saved from damage that could be caused by overload.
 

9. Imagine that you have three resistors of R Ω each. How many resultant resistances can be obtained by connecting these three in different ways? Draw the relevant diagrams.    (AS 5) 4 Marks
A: We can connect these three resistor in four ways.
i) Connecting them in series.

 

10. Explain the effects of electric shock on human body?      (AS 1) (AS 6) 4 Marks
A: The current passing through our body when we touch alive wire of 240 V is given by I = 0.0024 A. When this quantity of current flows through the body the functioning of organs inside the body gets disturbed. This disturbance inside the body is felt as electric shock.
        If the current flows continues further it damages the tissues of the body. which leads to decrease in resistance of the body when this current flows for a longer time damage to the tissues increases and thereby the resistance of human body decreases further. Hence, the current through the human body will increase. If this current reaches 0.07 A, it effects the functioning of the heart and if this much current passes through the heart for more than one second it could be fatal. If this current flows for a longer time, the person in electric shock is being killed.

EFFECTS OF THE ELECTRIC CURRENT ON HUMAN BODY:

Current in ampere	Effect
0.001	can be felt
0.005	is painful
0.010	Causes involuntary muscle contractions (Spasms)
0.015	Causes loss of muscle control
0.070	If through the heart, causes serious disruption, probably fatal if current laster for more than 1 sec.

11. Explain the sign convention used in electric circuit as per Loop's law.   (AS 1) 4 Marks
A: i) EMF of the battery is taken as negative when we move from positive terminal to negative terminal across the battery.
                                         

ii) It is taken as positive when we move across the battery from negative terminal to positive terminal.
                                        
iii) The direction of electric current is to be observed to give a sign to potential difference of a resistor.
iv) The potential difference across the resistor is taken as negative when we move along the direction of electric current through the resistor.
                                       
v) It is taken as positive when we move against the direction of electric current through the resistor.
                                        

12. Draw the circuit diagrams of all possible combinations with resistors R₁, R₂, R₃, R₄      (AS 5) 4 Marks

13. Find the resultant potential difference of the following loops from the given circuit diagram.   (AS 4) 4 Marks
    

    i) Loop ABEFA
    ii) Loop ACDFA
    iii) Loop BCDEB
    iv) Loop EFABE
    v) Loop DEBCD
    vi) Loop DFACD
A: i) For the Loop ABEFA:
   The resultant potential difference in the loop ABEFA is
    -V₂ + I₂R₂ - I₁R₁ + V₁ = 0

ii) For the Loop ACDFA:
The resultant potential difference in the loop ACDFA is
-(I₁ + I₂)R₃ - I₁R₁ + V₁ = 0
iii) For the Loop BCDEB
The resultant potential difference in the loop BCDEB is
-(I₁ + I₂)R₃ - I₂R₂ + V₂ = 0
iv) For the Loop EFABE:
The resultant potential difference in the loop EFABE is
-I₁R₁ + V₁ - V₂ + I₂R₂ = 0
v) For the Loop DEBCD:
The resultant potential difference in the loop DEBCD is
-I₂R₂ + V₂ - (I₁ + I₂)R₃ = 0
vi) For the Loop DFACD
The resultant potential difference in the loop DFACD is
-I₁R₁ + V₁ - (I₁ + I₂)R₃ = 0

14. Madhu drew a graph by taking flow of current on Y - axis and voltage on X - axis. He got the values by using a wire, voltmeter and ammeter?

Answer the following questions based on the graph.
i) What kind of wire was it?
ii) Find the resistance of the wire?
iii) How much of electrical energy is used by the wire when a potential difference of 20 V is supplied between the ends of the wire?
iv) Which rule/principle does the above graph show? (AS 4) 4 Marks
A: i) Graph is a straight line. So, the wire is a ohmic conductor (metal)
ii) From graph I = 0.1 A and V = 5 V
From Ohm's law V = IR
^{R =} V/I= %/0.1^{= 50 Ω}

iii) Potential difference between the ends of wire V = 20 V, R = 50 Ω 
     
iv) From above graph V/I is a constant ratio, hence the graph shows Ohm's law.
 

15. The electric circuit is shown in figure.
        
Find out the equivalent resistance between A and B?   (AS 1) 2 Marks
A: The first three resistors are in parallel.

16. Find the equivalent resistance between the points A and B in the given electric circuit?   (AS 1) 2 Marks
              
A: The voltage divides in three resistors if we connect a battery in between A and B. If we redraw the circuit,

17. Draw the picture which shows the connections of electrical appliances in your home.    (AS 5) 2 Marks
A:
                               

18. In household circuits, why do we use fuses?
                 (OR)
       What is the use of fuses?   (AS 6) 2 Marks
A: * The fuse consists of a thin wire of low melting point.
* When the current in the fuse exceeds 20 A, the wire will heat up and melt.
* The circuit then becomes open and prevents the flow of current into the household circuit.
* Hence all the electric devices are saved from damage that could be caused by overload.
* Thus we can save the house holding wiring and devices by using fuses.
 

19. Assume that the resistance of your body is 1,00,000. Ω If you touch a battery of 24 V. Then find the current flowing in your body. What will be the effect of that current on your body?  (AS 1) 2 Marks
A: Resistance of the body (R) = 1,00,000 Ω
Potential difference of the battery (V) = 24 V
Current that flows in the body (I) = ?

                                            I = 0.00024 A
       This current is very less quantity. When such less quantity current passes through the human body it does not affect the functioning of various organs inside the body.
 

20. Why doesn't a bird get a shock when it stands on a high voltage wire? (AS 6) 2 Marks
A: There are two parallel transmission lines on electric poles. The potential difference between these two lines is 240 V through out their lengths. If you connect any conducting device across these two wires, it causes current to flow between the wires. When the bird stands on a high voltage wire, there is no potential difference between the legs of the bird. Because it stands on a single wire. So, no current passes through the bird. Hence, it doesn't feel any electric shock.
 

21. Give reasons for using Lead in making fuses.  (AS 1) (AS 6) 2 Marks
A: Lead is used in making fuses. Because it have low melting point. If the current in the Lead wire exceeds certain value the wire will heat up and melts. So, the circuit in households opened and all the electric devices are saved.

22. Draw the symbols of the following.
i) Battery    ii) Resistance    iii) Ammeter     iv) Key    (AS 5) 2 marks

 

23. Draw the shape of V - I graph for a conductor.    (AS 5) 1 Mark

24. Draw the shape of V - I graph for a semi conductor.   (AS 5) 1 Mark
A: V - I graph for semi conductor:
           
 

25. What are the limitations of Ohm's law?    (AS 1) 1 Mark
A: Ohm's law is valid for metal conductors, provided the temperature and other physical conditions remain constant.
Ohm's law is not applicable to gaseous conductors.
Ohm's law is not applicable to semi conductors also such as Germanium and Silicon.
 

26. What are the uses of Semi conductors?   (AS 6) 1 Mark
A: Semi conductors are used to make diodes, transistors and integrated circuits (IC's). IC's are used in all sorts of electronic devices, including Computer, TV, Mobile, Phone etc.

27. Which instrument is used to measure electric current?  (AS 6) 1 Mark
A: Ammeter
 

28. Which instrument is used to measure potential difference?     (AS 1) 1 Mark
A: Voltmeter
 

29. Find the value of 'X' in the following figure?    (AS 1) 1 Mark
                       
A: From Junction law    I₁ + I₂ = I₃ + I₄
                                           2 + 5   = 1.5 + x
                                                    x = 7 - 1.5
                                                    x = 5.5 A

30. Find the resultant potential difference from the given figure based on Loops law.   (AS 4) 2 Marks
             
A: i) For the Loop ABCDA:
The resultant potential difference in the loop is
   +V₁ - IR₁ + V₂ - IR₂ - IR₃ = 0
ii) For the Loop ADCBA:
The resultant potential difference in the loop is
   +IR₃ + IR₂ - V₂ + IR₁ - V₁ = 0

PROBLEMS - SOLUTIONS

1. A total charge of 90 coulombs flows in a conductor during a time of 4 minutes. What is the strength of current in the conductor?  (AS 1) 2 Marks

Sol: Given Q = 90 Coulombs
t = 4 × 60 = 240 sec
I = ?
^{I = Q/t 
I =} 90/240
= 0.375 Amperes
 

2. What is the total quantity of charge flows in 5 minutes when a current of 2 amperes exist in a conductor?  (AS 1) 2 Marks
Sol: Given I = 2 Amperes
t = 5 × 60 = 300 sec
Q = ?
^{I = Q/t} 

Q = I × t
= 2 × 300
= 600 Coloumbs
 

3. The potential difference across a bulb is 240 V and a current of 6 A flows through it. Find the resistance of the bulb.  (AS 1) 2 Marks
Sol: Given V = 240 V
I = 6 A
R = ?
According to Ohm's Law
^{R = V/I}

^{R = 240/6}
= 40 Ω

4. An immersion heater of resistance 23 Ω is connected to a mains of 230 V supply. How much current flows through it?  (AS 1) 2 Marks
Sol: Given R = 23 Ω
V = 230 V
I = ? According to Ohm's Law 
^{I =} V/R
^{I =} 230/23
   = 10 A
 

5. 0.15 A current is flowing through a resistance of 10 Ω. What is the potential difference between its ends?    (AS 1) 2 Marks
Sol: Given R = 10 Ω
I = 0.15 A V = ?
According to Ohm's Law
V = IR
V = 0.15 × 10
= 1.5 V

6. Assume that the resistance of your body is 5,00,000 Ω and if you touch a 240 V wire, then how much current flows through your body.  (AS 1) 2 Marks
Sol: Given R = 5,00,000 Ω
V = 240 
I = ?
^{According to Ohm's Law I = V/R
I =} 240/5,00,000
= 0.00048 A
 

7. The resistance of a manganin wire of 1 m length is 8 Ω. Find the resistance of a wire of 5 m length of same material having the same area of cross-section.  (AS 1) 2 Marks
Sol: Given R₁ = 8 Ω
l₁ = 1 m
l₂ = 5 m
R₂ = ?

Since R ∝ l, we can write

R₂ = 40 Ω
 

8. The resistance of a manganin wire of 1 mm² cross-sectional area is 12 Ω.
Find the resistance of manganin wire of same length. But of a cross-section of 4 mm2.   (AS 1) 2 Marks
Sol: Given R1 = 12 Ω
A1 = 1 mm²
A² = 4 mm²
R² = ? 

 

9. The resistance of a brass wire of length 300 m and area of cross-section 3.4 × 10^-6 m² is 6 Ohms. Find the specific resistance of the material of the wire. (AS 1) 2 Marks
Sol: Given l = 300 m
A = 3.4 × 10^-6 m²
R = 6 Ω
ρ = ?
Specific resistance

ρ = 6.8 × 10^-8 Ohm - meter
 

10. What is the equivalent resistance of two resistors 4 Ω and 8 Ω when connected in (a) series and (b) parallel. (AS 1) 4 Marks
Sol: Given R₁ = 4 Ω
                     R₂ = 8 Ω
(a) When resistance are connected in series the equivalent resistance Req is given by
R_eq = R₁ + R₂
R_eq = 4 + 8
R_eq = 12 Ω
(b) When resistance are connected in parallel the equivalent resistance is given by

 

11. Calculate the equivalent resistance of two resistors of 1 Ω and 10 Ω connected in parallel.  (AS 1) 2 Marks
Sol: Given R₁ = 1 Ω
                    R₂ = 10 Ω
When resistance are connected in parallel the effective resistance R is given by

12. Three resistors 2 Ω, 4 Ω, 8 Ω are connected in (a) series, (b) parallel. Find the resultant resistance in the circuit.   (AS 1) 4 Marks
Sol: Given R₁ = 2 Ω
                     R₂ = 4 Ω
                     R₃ = 8 Ω
(a) Resultant resistance in series connection
R_eq = R₁ + R₂ + R₃
R_eq = 2 + 4 + 8
R_eq = 14 Ω
(b) Resultant resistance in parallel connection

13. Three resistors R1 Ω, 2 Ω and 8 Ω are connected in series in a circuit. If the resultant resistance in the circuit is 17 Ω. Find the value of R1.     (AS 1) 2 Marks
Sol: Resultant resistance in series connection
Req = R₁ + R₂ + R₃
17 = R₁ + 2 + 8
R₁ = 17 - 10
.^.. R₁ = 7 Ω

14. The effective resistance of a combination of three resistors is 50 Ohms. The value of two resistors are 10 Ω and 15 Ω. Find the value of third resistance?  (AS 1) 2 Marks
Sol: Given R_eq = 50 Ω
R₁ = 10 Ω
R₂ = 15 Ω
R₃ = ?

Resultant resistance in series connection
R_eq = R₁ + R₂ + R₃
50 = 10 + 15 + R₃
R₃ = 50 - 25
.^.. R₃ = 25 Ω
 

15. Two resistors are 6 Ω and R2 Ω are connected in parallel in a circuit. If the resultant resistance in the circuit is 4 Ω. Find the value of R2.    (AS 1) 2 Marks
Sol: Resultant resistance in parallel connection

 R₂ = 12 Ω

16. Find the resultant resistance in the circuit between A and B from the below diagram.
                        
Sol. Given R₁ = 3 Ω
R₂ = 6 Ω
R₃ = 2 Ω
From the given circuit
R₁ = 3 Ω, R₂ = -6 Ω are connected in parallel their resultant resistance is
 
.^.. R_eq = 2 Ω
But R₁, R₂ are connected to R₃ in series their resultant resistance is
R_eq = 2 + 2 = 4 Ω

 
From the above circuit diagram find the effective resistance between X and Y.  (AS 1) 4 Marks
Sol. Resistance of I parallel combination is