ELECTRICITY - IV : NUMERICALS

SOLVED NUMERICALS OF TEXT BOOK (GSEB)

*INDEX-TOPIC SEARCH

1. If an electric lamp lights for 2 h drawing 0.4 A of current, calculate the amount of charge that has passed through the lamp.
   
   SOLUTION
   
   Here
   
   Current I = 0.4 A
   
   Time t = 2 h = 2 x 3600 = 7200 seconds
   
   Charge Q = ?
   
   Now, Q = I x t
   
   ∴ Q = 0.4 x 7200 = 2880 = 2.88 x 10³ C        (ANS)



2. How much current will an electric heater having resistance 45 ohm draw if it is connected to a 220 V line ?
   
   SOLUTION
   
   Here
   
   Resistance R = 45 Ω
   
   Voltage V = 220 V
   
   Current I = ?
   
   According to Ohm's law,
   
   I = V/R
   
   ∴ I = 220/45 = 4.89 A        (ANS)
   



3. A lamp of resistance 20 Ω is joined to a battery of 12 V. Find the value of the resistance to be connected in series with the lamp in order to get current of 0.5 A.
   
   SOLUTION
   
   
   
   Here
   
   Voltage V = 12 V
   
   Current I = 0.5 A
   
   Resistance of the lamp R₁ = 20 Ω
   
   Resistance to be connected R₂ = ?
   
   If R is the equivalent resistance of the circuit, then
   
   R = V/I = 12/ 0.5 = 24 Ω
   
   But,       R = R₁ + R₂
   
          ∴ 24 = 20 + R₂
   
          ∴ R₂ = 4 Ω        (ANS)
   



4. Calculate the net(equivalent) resistance of the circuit if resistances of 10 Ω, 5 Ω and 15 Ω are connected in parallel.
   
   SOLUTION
   
   Let the three resistances connected in parallel be
   
   R₁ = 10 Ω, R₂ = 5 Ω, R₃ = 15 Ω
   
   If their net resistance is R, then
   
           1/R = 1/R₁ + 1/R₂ + 1/R₃
   
           ∴ 1/R = (1/10) + (1/5) + (1/15)
   
           ∴ 1/R = (3 + 6 + 2)/30 = 11/30
   
           ∴ R = 30/11 = 2.73 Ω        (ANS)
   



5. Three resistances of 5 Ω, 10 Ω and 30 Ω are connected in parallel with a battery of 12 V. Find out(a)the current passing through each resistance, (b)the total current of the circuit,(c) the equivalent resistance of the circuit.
   
   SOLUTION
   
   
   
   Here the three resistances are
   
   R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 30 Ω
   
   Voltage V = 12 V
   
   Let the currents passing through the three resistances be I₁,I₂,I₃,respectively and the equivalent resistance be R.
   
   Using Ohm's law, we get
   
      I₁ = V/R₁
   
       ∴ I₁ = 12/5 = 2.4 A
   
      I₂ = V/R₂
   
       ∴ I₂ = 12/10 = 1.2 A
   
      I₃ = V/R₃
   
       ∴ I₃ = 12/30 = 0.4 A
   
   Thus, net current I = I₁ + I₂ + I₃
   
         ∴ I = 2.4 + 1.2 + 0.4 = 4.0 A
   
   Again, 1/R = 1/R₁ + 1/R₂ + 1/R₃
   
           ∴ 1/R = 1/5 + 1/10 + 1/30
   
           ∴ 1/R = 1/3
   
           ∴ R = 3 Ω       (ANS)



6. In the following figure if five resistors have been connected as shown,find the equivalent resistance and the net current of the circuit.Given R₁ = 10 Ω,R₂ = 40 Ω,R₃ = 30 Ω,R₄ = 20 Ω,R₅ = 60 Ω and voltage of the battery V = 12 V.
   
   
   
   SOLUTION
   
   Here voltage V = 12 V
   
   If the equivalent resistance of R₁ and R₂ is R' then
   
   1/R' = 1/R₁ + 1/R₂
   
   ∴ 1/R' = 1/10 + 1/40 = 0.1 + 0.025 = 0.125
   
   ∴ R' = 1/0.125 = 8 Ω
   
   If R" is the equivalent resistance of R₃, R₄ and R₅ then
   
   1/R" = 1/R₃ + 1/R₄ + 1/R₅
   
   ∴ 1/R" = 1/30 + 1/20 + 1/60 = 1/10
   
   ∴ R" = 10 Ω
   
   If R is the equivalent resistance of R' and R" then
   
   R = R' + R"        (R' and R" are in series)
   
   ∴ R = 8 + 10 = 18 Ω
   
   and current I = V/R = 12/18 = 0.67 A       (ANS)
   



7. If an electric lamp draws 500 mA with voltage of 2.5 V, calculate the power and the energy consumed in 1 minute.
   
   SOLUTION
   
   Here, voltage V = 2.5 V, current I = 500 mA = 0.5 A, time t = 1 min = 60 seconds
   
   Now, power P = V x I = 2.5 x 0.5 = 1.25 watt
   
   and energy consumed in 1 minute W = P x t
   
           ∴ W = 1.25 x 60 = 75 joule       (ANS)
   



8. A 100 W lamp glows for 2 hours per day calculate the energy consumed in 30 days.
   
   SOLUTION
   
   Here power of lamp P = 100 watt = 0.1 kW
   
   time t = 2 x 30 = 60 h       (2 h/day x 30 days)
   
   Now energy consumed in 30 days
   
   W = P x t = 0.1 x 60 = 6 kWh       (ANS)



9. 
   
   SOLUTION

UNSOLVED NUMERICALS OF TEXT BOOK (GSEB)

1. If 300 mA current passes through a lamp, how many electrons will pass through it in 1 minute ?
   
   SOLUTION:
   
   Here, current I = 300 mA = 0.3 A
   
      time t = 1 minute = 60 seconds
   
   Now, charge Q = I x t = 0.3 x 60 = 18 coulomb
   
   and no. of electrons = (total charge)/(charge on 1 electron)
   
       ∴ no. of electrons = (18)/(1.6 x 10^-19)
   
                       = 1.125 x 10²⁰ electrons        (ANS)
   


2. If a current of 5 mA passes through a resistance of 2200 Ω, find out the voltage drop on it.
   
   SOLUTION:
   
   Here, current I = 5 mA = 0.005 A and resistance R = 2200 Ω
   
   Now, voltage drop V = IR = 0.005 x 2200 = 11 V     (ANS)
   


3. Three resistors R₁ = 5 Ω, R₂ = 8 Ω and R₃ = 12 Ω are connected in series with a battery of 4 V. Find the value of current in the circuit and the voltage drop across each resistor.
   
   SOLUTION:
   
   
   Here,
   
   Equivalent resistance R = R₁ + R₂ + R₃
   
          ∴ R = 5 + 8 + 12 = 25 Ω
   
   Now, current I = V/R =4/25 = 0.16 A
   
   If V₁, V₂ and V₃ are voltage drops across R₁, R₂, and R₃,respectively, then
   
   V₁ = I x R₁ = 0.16 x 5 = 0.8 V
   
   V₂ = I x R₂ = 0.16 x 8 = 1.28 V
   
   V₃ = I x R₃ = 0.16 x 12 = 1.92 V        (ANS)
   


4. This numerical has been included in the previous one.
   
   SOLUTION:
   Requires no solution.
   


5. Three resistors are joined in parallel. A current of 7.5 A flows due to 30 V battery. If two resistors have resistances of 10 Ω and 12 Ω. Find out the third. Also find the current passing through each resistor.
   
   SOLUTION:
   
   
   
   Let the three resistances in parallel be : R₁ = 10 Ω, R₂ = 12 Ω and R₃.
   
   Here voltage V = 30 V and current I = 7.5 A
   
   ∴ Net resistance of the circuit R = V/I = 30/7.5 = 4 Ω
   
   Also, 1/R₁ + 1/R₂ + 1/R₃ = 1/R
   
   ∴ 1/10 + 1/12 + 1/R₃ = 1/4
   
   ∴ 1/R₃ = 1/4 - 1/10 - 1/12 = 1/15
   
   ∴ R₃ = 15 Ω
   
   If I₁, I₂ and I₃ are the currents passing through R₁, R₂ and R₃ respectively, then
   
   I₁ = V/R₁ = 30/10 = 3.0 A
   
   I₂ = V/R₂ = 30/12 = 2.5 A
   
   I₃ = V/R₃ = 30/15 = 2.0 A ___________[ANS]
   


6. Calculate the current flowing through the circuit shown below.
   
   
   
   SOLUTION:
   
   We can draw the circuit as shown below:
   
   
   
   Here R₁ and R₂ are connected in series , therefore, their equivalent resistance R' is given by
   
   R' = R₁ + R₂ = 30 + 30 = 60 Ω
   
   Again, R' and R₃ are in parallel,so their equivalent resistance R can be calculated by
   
   1/R = 1/R' + 1/R₃
   
   ∴ 1/R = 1/60 + 1/30 = 3/60 =1/20
   
   ∴ R = 20 Ω
   
   Thus the equivalent resistance of the circuit R = 20 Ω.
   
   Now, current I = V/R =2/20
   
           ∴ I = 0.1 A _________(ANS)
   


7. Two lamps of 100 W and 60 W are joined in parallel with a line voltage of 220 V. Calculate the current flowing through the circuit.
   
   SOLUTION:
   Here, the total power of the lamps P = 100 + 60 = 160 W
   
   Now, current I = P/V = 160/220 = 0.73 A ________(ANS)
   


8. An electric iron draws a current of 5.0 A. If its resistance is 44 Ω, calculate the energy consumed in 5 minutes.
   
   SOLUTION:
   
   Here, current I = 5 A, resistance r = 44 Ω,
   
   time t = 5 minutes = 5 x 60 = 300 seconds.
   
   Now, W = I² R t = (5)²(44)(300) = 330000 J_____(ANS)
   


9. Two lamps of 100 W and 500 W are joined in parallel with a main line. Which lamp will glow brighter ? Why ?
   
   SOLUTION:
   
   The lamp with 500 W will glow brighter because it is using five times the energy used by the 100 W lamp per second.(ANS)
   


10. In a house three lamps of 100 W, 60 W and 40 W are kept on for 25 hours. How many units of energy will be consumed ?
    
    SOLUTION:
    
    The total power of three lamps = 100 + 60 + 40 = 200 W
    
    ∴ Energy consumed in 25 hours = 200 x 25 = 5000 Wh
    
                    = 5 kWh = 5 unit______(ANS)
    


11. 
    
    SOLUTION: