Q: State Ohms law. Write the necessary conditions for its validity. How is this law verified experimentally? What will be the nature of graph between potential difference and current for a conductor? Name the physical quantity that can be obtained from this graph.

Ohm’s law : When the physical conditions such as temperature etc. remain same, the current flowing through the conductor is directly proportional to the potential difference applied across the ends of the conductor, i.e., I \propto V or V \propto I \Rightarrow \dfrac{V}{I} = constant \Rightarrow V = IR where R is constant of proportionality and is called resistance of the wire.Necessary condition for validity of Ohm's Law is that physical condition such as temperature of the conductor remains same. "Procedure" V-I graph is a straight line passing through the origin of the graph as shown. Conclusion : Straight line nature of graph shows that the current is proportional to the potential difference. Hence, Ohm’s law verified. The slope of V-I graph gives the value of resistance of the conductor at the given temperature.

Question 1

State Ohms law. Write the necessary conditions for its validity. How is this law verified experimentally? What will be the nature of graph between potential difference and current for a conductor? Name the physical quantity that can be obtained from this graph.

Accepted Answer

Ohm&#8217;s law : When the physical conditions such as temperature etc. remain same, the current flowing through the conductor is directly proportional to the potential difference applied across the ends of the conductor, i.e.,      I \propto V    or     V \propto I     \Rightarrow                  \dfrac{V}{I}   = constant   \Rightarrow                       V = IR   where  R  is constant of proportionality and is called resistance of the wire.Necessary condition for validity of Ohm's Law is that physical condition such as temperature of the conductor remains same. &#34;Procedure&#34;  V-I graph is a straight line passing through the origin of the graph as shown. Conclusion : Straight line nature of graph shows that the current is proportional to the potential difference. Hence, Ohm&#8217;s law verified. The slope of V-I graph gives the value of resistance of the conductor at the given temperature.

Question 2

The voltage - current  (V-I)  graph of a metallic conductor at two different temperature  T_{1}  and  T_{2}  is shown below. At which temperature is the resistance higher?

Accepted Answer

The mathematical expression for Ohm's law is:                                    V=IR Thus, graph between V versus I gives a straight line with slope R.Since, slope at  {T_{2}}  is more than slope at  {T_{1}} . For metallic conductors, resistance is directly proportional to temperature. Thus, resistance is higher at  {T_{2}}

Question 3

Graphs between electric current and potential difference across two conductors A and B are shown in the figure. Which of the following conductor has more resistance?

Accepted Answer

For comman voltage  V  Current in conductor B is greter then currecnt in conductor A.   ie     i_B>i_A          for same value of VFrom ohms law,    R=\dfrac VI\R\propto\dfrac1I Hence, If  i_B>i_A                R_A>R_B  Option B

Question 4

Three bulbs each having power P are connected in series in an electric circuit. In another circuit, another set of three bulbs of same power are connected in parallel to the same source.(i) Will the bulbs in both the circuits glow with the same brightness? Justify your answer(ii) Now let one bulb in each circuit get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.(iii) Representing each bulb by a resistor, draw circuit diagram for each case.

Accepted Answer

(i) Bulbs in parallel provide more illumination. This is because(a) each bulb gets same voltage and is equal to the applied voltage.(b) each bulb draws required current from the mains. Hence, they work properly.(ii)When one bulb in each circuit get fused, In series: Rest of the bulbs will not glow. This is because in series arrangement, there is only a single path for the flow of current. In parallel: Rest of the bulbs will continue to glow as in parallel connection,(a) individual branch in the circuit completes its own circuit, or(b) different paths are available for the flow of current.(iii) Circuit diagram

Question 5

In the circuit diagram find:(A) Equivalent resistance(B) Current shown by the ammeter A

Accepted Answer

Given: a circuit with three resistor in combination of parallel and series, a  4V  battery and an ammeter, A.To find (A) equivalent resistance, (B) current shown by the ammeter ASolution:Let,  R_1=5\Omega, R_2=3\Omega, R_3=2\Omega, V=4V And let the total current flowing in the circuit be  I From the circuit, it is clear that  R_2  and  R_3  are in series,Therefore,  R_s=R_2+R_3=3+2=5\Omega And this series combination is in parallel with  R_1 ,,Therefore, the equivalent resistance of the circuit is,  R_{eq}=\dfrac {R_1	imes R_s}{R_1+R_s}\\implies R_{eq}=\dfrac {5	imes 5}{5+5}\\implies R_{eq}=\dfrac {25}{10}=2.5\Omega Now Net current in the circuit is,  I=\dfrac VR=\dfrac 4{2.5}=1.6A So the ammeter reading is  1.6A

Question 6

In the circuit shown in the figure calculate the equivalent resistance between the points a and b.

Accepted Answer

{ R }_{ 2 }  and  { R }_{ 3 }  are in parallel then their equivalent is in series with  { R }_{ 1 } . { R }_{ eq }=\dfrac { { R }_{ 2 }{ R }_{ 3 } }{ { R }_{ 2 }+{ R }_{ 3 } } +{ R }_{ 1 }

Question 7

For the circuit given below, calculate the equivalent resistance between the points P and Q.

Accepted Answer

7,8,9 are in parallel \dfrac { 1 }{ { R }_{ { eq }_{ 1 } } } =\dfrac { 1 }{ 7 } +\dfrac { 1 }{ 8 } +\dfrac { 1 }{ 9 } \ { R }_{ { eq }_{ 1 } }=\dfrac { 7	imes 8+8	imes 9+7	imes 9 }{ 7	imes 8	imes 9 } \Omega \ =\dfrac { 191\Omega  }{ 504 }  5,4 are in parallel \dfrac { 1 }{ { R }_{ { eq }_{ 2 } } } =\dfrac { 1 }{ 5 } +\dfrac { 1 }{ 4 } \ { R }_{ { eq }_{ 2 } }=\dfrac { 20\Omega  }{ 9 }  { R }_{ { eq }_{ 1 } }  and  { R }_{ { eq }_{ 2 } }  are in seriesNet resistance  ={ R }_{ { eq }_{ 1 } }+{ R }_{ { eq }_{ 2 } }\ =\dfrac { 191 }{ 504 } +\dfrac { 20 }{ 9 } \ { R }_{ net }=\dfrac { 191+20	imes 56 }{ 504 } \quad \ { R }_{ net }=\dfrac { 191+1120 }{ 504 } \ { R }_{ net }=2.6\Omega

Question 8

(a) Define the term volt.(b) State the relation between work, charge and potential difference for an electric circuit.Calculate the potential difference between the two terminals of a battery if  100\ J  of work is required to transfer  20\ C  of charge from one terminal of the battery to the other.

Accepted Answer

(a) When  1  joule of work is done in carrying  1  coulomb of charge , from infinity to a point in the electric field, then potential at that point is called  1  volt. (b) Potential difference , V = Work done on unit charge = W/q Work is  100\ J , q = 20\ C   Potential difference ,  V = W/q = 100/20 = 5\ V

Question 9

A piece of wire of resistance  20  \Omega  is drawn out so that its length increases to twice its original length. Calculate the resistance of the wire in the new situation.

Accepted Answer

When length of wire is doubled i.e.  2l  Also it's area of cross section becomes half i.e  \dfrac A2 Resistance  R =\dfrac{ &#961;l}{A}= 20 \Omega New Resistance , R'=\dfrac{ho l'}{A'} =\dfrac{ &#961;(2l)	imes 2}{A} =4\dfrac{ &#961;l}{A}= 4(\dfrac{ &#961;l}{A}) = 4	imes 20 = 80\Omega

Question 10

An electric heater of resistance  6\ \Omega   is run for  10\ min  on a line of  120\ V . The energy liberated in this period of time is:

Accepted Answer

By Joule's Law of heating: E=\dfrac { { V }^{ 2 } }{ R } t E ; Energy developed V ; Applied voltage R ; Resistance of wire t ; time  \Rightarrow E=\dfrac { 120^2}{ 6 } 	imes 10	imes 60 \Rightarrow E=14.4	imes { 10 }^{ 5 }J

Question 11

A house is fitted with ten lamps of each  60\ W . Each lamp burns for  5\ hrs  a day on an average. The cost of consumption in a month of  30\ days  at  2. 80  rupees per unit is:

Accepted Answer

Total energy spent is given by:  E = nE'  where  n:  number of bulbs            E':  Energy spent by each bulb E' = Pt E' = 60 	imes 5 	imes 30\ Wh  E = 60 	imes 5 	imes 30 	imes 10         = 9000 	imes 10          = 90 KWh Cost  = 90 	imes 2.8              = 28 	imes 9             = 252 Rs

Question 12

There are  5  tube-lights each of  40\ W  in a house. These are used on an average for  5\ hrs/day . In addition, there is an immersion heater of  1500\ W  used on an average for  1\ hr/day . The number of units of electricity that are consumed in a month of  30\ days  is:

Accepted Answer

E=n_1{ P }_{ 1 }{ t }_{ 1 }+n_2{ P }_{ 2 }{ t }_{ 2 } E= 5  	imes 40	imes 5	imes 30+1 	imes 1500	imes 30      =200	imes 150+45000 E=75\ kWh \Rightarrow E = 75\ units

Electricity