Q: An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be :

\textbf{Step 1: Finding Resistance of bulb using given Ratings} Rating of bulb P=100W, V=220V Power, P=\dfrac{V^2}{R} \Rightarrow\ \ R=\dfrac{V^2}{P} =\dfrac{220\times 220}{100}=484 \Omega \textbf{Step 2: Power consumed by bulb } Now the voltage drop across the bulb is V'=110V So, power consumed by the bulb will be: P_b=\dfrac{V'^2}{R}=\dfrac{110\times 110}{484} \text{(Resistance of bulb will remain same)} P_b=25 W The correct option is D.

Q: Which one of the following is not the unit of energy?

\textbf{Hint:} Use the relation between energy and power. \textbf{Explanation :} If Watt is unit of power then kilowatt is also the unit of power.Work done = Force ×× Distance = Newton - metre, Therefore it is the unit of Work or Energy.Energy = Power ×× time = kilowatt - hour, in fact it is the commercial unit of Energy.

Q: A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly:

\textbf{Given : } I= 1A, t=16s, e= 1.6\times 10^{-19} C \textbf {Step 1: Current drawn by the filament} I= \dfrac{Q}{t} = \dfrac{Ne}{t} (Since Q= Ne , where N is the number of electrons crossing in time t) \Rightarrow N= \dfrac{I \ t}{e} ....(1) \textbf {Step 2: Substituting the values in equation (1)} \ N=\dfrac { I\ t }{ e } =\dfrac { 1A\times 16 s} { 1.6\times { 10 }^{ -19 } C} ={ 10 }^{ 20 } Hence, number of electrons passing through a cross section of the filament in 16 seconds would be { 10 }^{ 20 } .

Q: If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

\textbf{Step 1: Calculation of power } The new value of current I'=I + 100\% of I =2I Initial power, P=I^2 R Now, the new power, { P }^{ ' }={ { I }^{ ' } }^{ 2 }R\ =4{ I }^{ 2 }R\ = 4P \textbf{Step 2: Percentage increase in power } =\dfrac{{ P }^{ ' }-P}{P}\times 100 =\dfrac{4P-P}{P}\times 100=300\% Hence, the increase in power dissipated will be 300%.Hence, option C is correct.

Question 1

An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be :

Accepted Answer

extbf{Step 1: Finding Resistance of bulb using given Ratings}  Rating of bulb  P=100W,   V=220V Power,  P=\dfrac{V^2}{R} \Rightarrow\ \ R=\dfrac{V^2}{P}   =\dfrac{220	imes 220}{100}=484 \Omega 	extbf{Step 2: Power consumed by bulb } Now the voltage drop across the bulb is  V'=110V So, power consumed by the bulb will be: P_b=\dfrac{V'^2}{R}=\dfrac{110	imes 110}{484}                 	ext{(Resistance of bulb will remain same)} P_b=25 W The correct option is D.

Question 2

Charge on an electron is  1.6 	imes 10^{-19}  coulomb. Number of electrons passing through the wire per second on flowing of 1 ampere current through the wire will be

Accepted Answer

The current is defined as rate of flow of charged particle.i.e.i    =  \dfrac{Q}{t}  Hence the charge flown is,Q   = i \  t  Hence Q   = 1.0 	imes 1  = 1 C  The charge Q is due to electrons, therefore,Q   = N	imes e    Here Q   = 1 C  If N is number of electrons passing and e is the electronic charge of value   1.6 	imes {10}^ {-19}  Then,N   = \dfrac {Q}{e} = \dfrac {1}{1.6 	imes {10}^ {-19} } = 0.625  	imes {10}^ {19} electrons

Question 3

State joule's law of heating.

Accepted Answer

Joule’s law of heating states that,when a current ‘i '
passes through a conductor of resistance ‘r’ for time ‘t’ then the heat
developed in the conductor is equal to the product of the square of the
current, the resistance and time.

Question 4

Which one of the following is not the unit of energy?

Accepted Answer

extbf{Hint:}  Use the relation between energy and power. 	extbf{Explanation :} If Watt is unit of power then kilowatt is also the unit of power.Work done = Force ×× Distance = Newton - metre, Therefore it is the unit of Work or Energy.Energy = Power ×× time = kilowatt - hour, in fact it is the commercial unit of Energy.

Question 5

In an electrical circuit two resistors of 2  \Omega  and 4  \Omega   respectively are connected in series to a 6 V battery. The heat dissipated by the 4   \Omega   resistor in 5 s will be

Accepted Answer

Hint : Use the mathematical expression of Joule's Law to evaluate heat dissipated.Step 1: Find the equivalent resistance of the Circuit.                    Given: Two resistors  R_{1}  &   R_{2}  are connected in series.                     So, equivalent resistance can be given as:                     R_{net}  =  R_{1}\ + \  R_{2}                       R_{net}  =  2\ + \  4 =   6 \Omega Step 2 : Find the current through  4 \Omega  resistor                     Since both resistors are connected in series so current through                      each resistor is same as the current through the circuit.                      Current = \dfrac{P.D.}{Net\ resistance}                        I =\dfrac{6}{6}=1A  Step 3: Heat dissipated through  4 \Omega  resistor can be given as :                        H= I^2Rt                        where,  H =  Heat dissipated,  R = Resistance and  t =  time in seconds                        So,  H = 1^2×4×5=20J Final Step : Heat dissipated through  4 \Omega  resistor in  5\ sec  is   20\ J . Option C is correct.

Question 6

An electric bulb of  60\ W  is used for  6\ hrs/day . Calculate the units of energy consumed in one day by the bulb.

Accepted Answer

Given :     t = 6\ hrs                               P = 0.06\ kW Thus energy consumed per day is: E=  Pt = 0.06	imes 6  = 0.36\ kWh  = 0.36\ 	ext{units}

Question 7

An electric lamp of resistance  20 \, \Omega  and a conductor of resistance  4\, \Omega  are connected to a  6\, V  battery as shown in the circuit. Calculate:(a) The total resistance of the circuit(b) The current through the circuit(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and (d) Power of the lamp.

Accepted Answer

a). the total resistance of the circuit = 20\Omega +4 \Omega = 24\Omega  b). The current through the circuit= current through the bulb= current through the conductor.The current through the circuit=  \dfrac{voltage\  applied}{Total\ Resistance}. the current through the circuit  =\dfrac{6}{24}amp The current through the circuit =  0.25amp c). The potential difference across the lamp and the conductor.(i) The potential difference across the electrical lamp  =0.25	imes 20= 5 volt (ii) The potential difference conductor  =0.25 	imes 4= 1 \ volt d).  Power can be calculated as  \dfrac{voltage \ across\ the \ lamp}{current\  through\  the\  lamp}      power= \dfrac{5}{0.25}ightarrow 20 W .

Question 8

Draw the symbols of the following circuit components.i) Electric cellii) Switch in off positioniii) Electric bulbiv) Battery

Accepted Answer

The symbols of circuit components are:

Question 9

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly:

Accepted Answer

extbf{Given : } I= 1A,  t=16s,  e= 1.6	imes 10^{-19} C 	extbf {Step 1: Current drawn by the filament}    I= \dfrac{Q}{t} = \dfrac{Ne}{t}          (Since  Q= Ne , where N is the number of electrons crossing in time t) \Rightarrow N= \dfrac{I \ t}{e}                                                                                                                                             ....(1) 	extbf {Step 2: Substituting the values in equation (1)}  \ N=\dfrac { I\ t }{ e } =\dfrac { 1A	imes 16 s} { 1.6	imes { 10 }^{ -19 } C} ={ 10 }^{ 20 } Hence, number of electrons passing through a cross section of the filament in 16 seconds would be  { 10 }^{ 20 } .

Question 10

If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

Accepted Answer

extbf{Step 1: Calculation of power } The new value of current  I'=I + 100\%   of  I   =2I Initial power,   P=I^2 R Now, the new power,  { P }^{ ' }={ { I }^{ ' } }^{ 2 }R\ =4{ I }^{ 2 }R\  = 4P 	extbf{Step 2: Percentage increase in power }                     =\dfrac{{ P }^{ ' }-P}{P}	imes 100                                       =\dfrac{4P-P}{P}	imes 100=300\% Hence, the increase in power dissipated will be 300%.Hence, option C is correct.