Solved Numericals of Physics Class 10 Electricity

 In this post, you will find solved numericals of Physics class 10 electricity based on work, power, and energy consumption by appliances. These numericals of Chapter 12 electricity are important for the coming Board Exams.

Numerical Problems Based on Work, Power, and Energy

The following formulas can be used to solve the numericals based on the work, power, and energy.

     `V=  W/Q`  [Note = We can put E in place of W ]

      `W=V×Q`

      `W=I^2 Rt` 

      `W=(V^2 t)/R`

      `P=  W/t`

      `P=  V^2/R`

       `P=I^2 R` 

   Where 

  P= power – Watt

  W = work – Joule

  V= voltage – Volt

  I  = current – Ampere

  Q = charge = Coulomb

   Q1. An electric bulb is connected to 110 volt electric source. If the current is 0.5 A, then what is the power of the bulb?

      Given values

    Voltage (V) = 110 volt

    Current (I) = 0.5 A

    Power (P)= ?

   Solution

    P=VI

    P =110 ×0.5

    P= 55 Watt

    Q2. A bulb rated 5 volt – 100mA, calculate its (i) power (ii) resistance

   Given values

   Voltage (V) =5 volt

   Current (I) = 100mA =  100 ×10^-3 A

   Solution

   P=VI

   P = 5 × 100 ×10^-3

   P = 0.5 Watt

    Q3. Two electric bulbs rated 60W,220V and 100W, 220 V. Which one of them has higher resistance?

   Given values

   (i)      P₁ = 60 W, V= 220 volt

   (ii)  P₂ = 110 W , 220 volt

         R₁=?

         R₂ =?

    Solution 

     `R_1=V^2/P_1` 

     `R_1=(220)^2/60`

    `R_1=806.67` Ohm

 

    `R_2=V^2/P_2` 

     `R_2= (220)^2/100`

     `R_2=484` Ohm

60-watt bulb has higher resistance than 100 watt bulb.

    Q4. Two lamps rated 100 W,220 Volt, and 60W, 220 V connected in parallel. Calculate the current drawn by the circuit.

       Given values

    P₁= 100W

    P₂= 60 W

    V = 220volt

    I=?

   Solution 

    `I=V/R`

In the above formula, we don’t have the value of 'R' so we will have to find the value of 'R' 

    `R_1=V^2/P_1` 

   ` R_1=(220)^2/100` Ohm

    `R_2=V^2/P_2` 

    `R_2=(220)^2/60` Ohm

Both the lamps are in parallel, so the resultant resistance is 

    `1/R=  1/R_1 +1/R_2` 

    `1/R=  100/(220 × 220)+60/(220 ×220)`

    `1/R=  (100+60 )/(220 × 220)`

    `1/R=  160/(220 × 220)` Ohm

  Now we can find the value of current (I)

   `I=220/((220  ×220)/160)`

    `I=(220 ×160)/(220 ×220 )`

   `I=160/(220 ) = 0.72A`

    Q5. An electric lamp is rated 100W,220V. it is used for 10 hours daily then calculate energy consumed in KWh per day.

   Given values

   P= 100 W

   V= 220 Volt

   Solution 

 Energy consumed per day =  `(P ×t)/1000`

              =  `(100 ×10)/1000`

 Energy consumed per day = 1KWh or 1 Unit

    Q6. A bulb rated 2.5 V and 650mA. Calculate (i) power (ii) resistance (iii) energy consumed for 5 hours.

   Given values

   V=  2.5Volt

   I = 650 mA= 0.65A

   t = 5 hours

    Solution 

(iii). Energy consumed in 5 hours =   `(P ×t)/1000`

                                   

         =  `frac\{1.625 ×5}{1000} =0.008125`KWh

   Q7. An immersion rod of 750W is used for one hour. Find the energy consumed in (i) KWh (ii) Joule

       Given values

   P = 750 W

   t = 1 hour

       Solution 

     (i)  Energy consumed in KWh

     Energy consumed for 1 hour = `(P ×t)/1000`

        =  `(750 ×1)/1000=0.75` KWh

     (ii). Energy consumed in Joule

       1 KWh = `3.6 × 10^6`Joule

       So    `0.75 × 3.6 × 10^6`Joule

         = `2.7 × 10^6` Joule

   

    Q8. One heater is rated 220 V- 5 A. Calculate the energy consumed if it is used for 5 hours.

     

    Given values

     V= 220 volt

      I = 5 A

     Solution 

    Energy consumed for 5 hour =  `(P ×t)/1000`

     But in this formula, we need the value of Power (P)

       P= VI

       P = 220 ×5 

       P=1100W

    Now we can put the value of P in the formula 

    Energy consumed for 5 hour = `(1100 ×5)/1000`

      

                                      = 5.5 KWh

    Q9. Which uses more energy a 250W computer in 1 hour or a 1200W heater in 10 minutes.

   

      Given values

      P₁ = 250W

      t₁ = 1hour

      P₂ = 1200W

      t₂ = 10 min = `frac\{10}{60}` hour

        Solution

      (i)   A 250 W computer  is used for 1 hour

                                         = P₁ × t₁ = 250 ×1

                                        = 250Wh

     (ii)   A 1200W heater is used for 10 min

                                           = P₂ × t₂ = 1200 ×10/60

                                           = 200Wh

       So, the computer uses more energy.

  

   Q10. An electric heater is rated 1kW,220V. calculate its resistance.

        Given values

        P= 1kW= 1000W

        V= 220 volt

        R=?

    

      Solution 

        `R=V^2/P`

    

        `R=220^2/1000`

         R=48.4 Ohm

   Q11. An electric bulb rated 220 V- 100 W. Calculate its resistance.

      Given values

      Power(P) =100W

      Voltage (V)= 220volt

      Resistance (R) =?

   

       Solution 

       `P=  V^2/R`

      `100=  220^2/R`

      `R=  220^2/100`

       R= 484  Ohm

     ✅Related Topics

    1.  Derive the formula of work done when current flows through a conductor or resistor

    2. Numerial Based on Electric Current 

   3. Numericals Bases on Heating Effect of Electric Current

    4. 50 Numericals on Electricity pdf