Q 1 :    

An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is              [2024]

  • 12.5 W

     

  • 25 W

     

  • 100 W

     

  • 50 W

     

(A)     Resistance, R=V2P=(200)250=800Ω

          Power dissipated, P=(Vapplied)2R=(100)2800=1008=12.5W

 



Q 2 :    

The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is (Given e=1.6×10-19C)                              [2024]

  • 1.25×1019

     

  • 6.25×1017

     

  • 6.25×1018

     

  • 31.25×1017

     

(D)      Given: P = 110 W, V = 220 V

           Power, P=V.I110=(220)(I)I=0.5A

           I=qt=net0.5=(nt)×1.6×10-19

          (nt)=0.51.6×10-19=31.25×1017

 



Q 3 :    

Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _________ to ________ times of its initial length if the water is to be boiled in 15 minutes.                           [2024]

  • increased, 4/3

     

  • increased, 3/4

     

  • decreased, 3/4

     

  • decreased, 4/3

     

(C)     P=V2R,R=ρLAP=V2AρL or P1L

          P1×t1=P2×t2

          P1P2=t2t1=L2L1=1520L2=34L1

 



Q 4 :    

By what percentage will the illumination of the lamp decrease if the current drops by 20% ?                        [2024]

  • 46%

     

  • 26%

     

  • 36%

     

  • 56%

     

(C)       Pinitial=I2initialR

            Pfinal=(0.8 Iinitial)2R

            % change in power

           Pfinal-PinitialPinitial×100=(0.64-1)×100=-36%

           

 



Q 5 :    

The ratio of heat dissipated per second through the resistance 5Ω and 10Ω in the circuit given below is            [2024]

  • 4 : 1

     

  • 1 : 1

     

  • 2 : 1

     

  • 1 : 2

     

(3)

As parallel combination so, p.d. is same P=V2RP1R

P1P2=R2R1=105=21



Q 6 :    

When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become          [2024]

  • 1/4 W

     

  • 1/2 W

     

  • 2W

     

  • 4W

     

(4)

W=V2R                          ...(1)

and R=ρA

Now 

Req=R4

W'=V2R4=4V2R=4W



Q 7 :    

A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 Ω and a resistance R, to a 100 V mains as shown in figure. For the heater to operate at 62.5 W, the value of R should be _____ Ω.               [2024]



(5)

Rheater=V2P=(100)21000=10Ω

For heater P=V2RV=PR

V=62.5×10=25 V

i1=7510=7.5 A,  iH=2510=2.5 A

iR=i1-iH=7.5-2.5=5 A

V=IRR=255=5Ω



Q 8 :    

In the following circuit, the battery has an emf of 2V and an internal resistance of 23Ω. The power consumption in the entire circuit is ______ W.      [2024]



(3)

Req=43Ω

P=V2Req=443=3W