Numericals On Electricity Class 10

 In  this post, we will discuss numericals on current. These numerical questions are asked in the exams, so you must learn how to solve them.

Numericals On Electricity Class 10

Formula for Numerical Problems based on Electric Current

 `I=  Q/t`

 `I=  frac{ n e}{t}`

Where 

I  = Electric current – Ampere

Q = charge – Coulomb

t =  time  - second

n =  number of electrons 

e =  charge on an electron = `1.6 ×10^-19` C

Q1.The filament of bulb draws a current of 0.5 ampere. Calculate the amount of charge if bulb glows for 2.5 hrs.

  Given values 

 Electric current (I)= 0.5 A

 Time (t) = 2.5 Hrs = 2.5× 60 ×60 = 9000 sec

 Charge (Q) =?

Solution 

  `I=  Q/t`

  `Q = It`

  `Q = 0.5 ×9000` 

  `Q = 4500` Coulomb

Q2.10 Coulombs charge passing through a point in a circuit in 5 seconds. Find the electric current flowing in the circuit.

Given values 

Time (t)  = 5 sec

Charge (Q) = 10 C 

Electric current (I) =  ?

Solution 

 `I=  Q/t`

 `I=  10/5`

`I= 2 Amp.` 

Q3.A current of 0.5 Ampere is drawn by a filament of an electric bulb for 15 minutes. Find he amount of electric charge.

Given values 

Electric current (I)= 0.5  A

Time (t) = 15 min = 15× 60 = 900 sec

Charge (Q) =? 

solution

 `I=  Q/t`

`Q = I × t`

`Q = 0.5 × 900`

`Q = 450` Coulomb

Q4.A current of 1 Ampere is drawn by a filament of an electric bulb. Find the Number of electrons passing through a cross-section of the filament in 10 seconds.

Given values 

Electric current (I) = 1  A

Time (t) =15 s

Number of electrons (n) = ?

Charge on electrons (e) = `1.6 ×10^-19`  C

Solution 

 `I=  frac{n e }{t}`

After cross multiplication, we find 

 `n=  (I×t)/e`

Now put the given values in the formula

 `n=  (1 ×10)/(1.6 × 10^(-19) )`

` n=  100/(16 × 10^(-19) )`

  ` n=  6.25/( 10^(-19) )`

   `n= 6.25× 10^19`

Number of electrons (n) has no unit required.

Q5. A 150000 coulomb charge is required to deposit one mole of copper from the CuSO4 solution. Find the time to deposit 0.2 mole copper from the copper solution if a current of 5 Ampere is flowing through the solution.

Given values 

Electric current (I) = 5 Amp

Charge (Q) = 150000 C

Time (t) =?

Solution 

Charge required for 1 mole copper = 150000 C

Charge required for 0.2 mole copper = 150000 ×0.2 = 30000 C

 `I=  Q/t`

 `t=  Q/I =  30000/5`

 `t= 6000`s

Q6.An heater draws a current of 10 A for  5 minutes. Calculate the charge flowing in the heater.

Given values 

Electric current (I) = 10 A

Time (t) = 5 min = 5 × 60  =300 s

Charge (Q) = ?

Solutions

 `I=  Q/t`

 `Q=I×t`

 `Q=10 ×300`

  `Q=3000`   C

Q7.`1.25 ×10^18`  electrons are passed from one end to another end of a conductor in 10 seconds. Find the current flowing through the conductor.

Given values

Number of electrons (n) =  `1.25 ×10^18`  

Time (t) =10 s

Electric current (I) =?

Solution 

 `I= frac\{ n e}{t}`

[e = `1.6 ×10^-19` C]

 `I=  (1.25 ×10^(18 )  ×1.6 × 10^(-19)  )/10`

 I=0.2 A

Q8.Calculate the number of electrons present in 0.1 coulomb of charge.

Given values 

Charge (Q) = 0.1

Number of electrons (n) =?

Solution 

The number of elections in 1 Coulomb is `6.25 ×10^18`

The number of electrons in 0.1 Coulomb charge is = `6.25 ×10^18× 0.1`

     = `6.25 ×10^17`

  Numerical Problem Solution Video

Numerical Problems Based on Potential difference and Work

To solve the numericals on electricity, we will use the following formulas

 `V=W/Q`

Where

V= Potential difference – volt

W = work done  - Joule

Q = charge – Coulomb

Q9.Calculate the amount of work done in carrying an 8C charge from a terminal of 150 volts to 200 volts.

Given values 

Charge (Q) = 8 C

Potential (`V_1`) = 150 volt

Potential (`V_2`) = 200 volt

Potential difference (V) = `V_2 -V_1` 

                             200 – 150 = 50 volt

Work (W) =?

Solution 

  `V=W/Q`

 `W=VQ`

 `W=50× 8=400 J`

Q10. How much energy is given to each coulomb of charge passing through a 10 Volt battery?

Given values 

Potential difference/voltage (V) = 10 volt

Charge (Q) = 1 C

Energy (E) =?

Solution 

 `V=W/Q`

Work is defined as energy transferred by force, so 

We can put 'E' in the place of  'W'

 

 `E= VQ`

  = 10 ×1= 10 J

Q11. Work done in moving 5 coulomb charge from point A to B is 80 Joule, if potential at point A is 25 volt then find the potential at B.

Given values 

Work (W) =80 J

Charge (Q) = 5 C

Potential (`V_A`) = 25 volt

Potential (`V_B`) =?

Solution 

`V=V_B   - V_A =  W/Q` 

 `V_(B  )-25=  80/5`

 `V_(B   )-25= 16`

 `V_(B   )= 16+25`= 41 volt

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