 # Electricity class 10 numericals

Last updated on May 31st, 2023 at 04:55 pm

In this post, we will solve a set of Electricity class 10 Numericals. These numerical problems with the solution will be a real help for the students preparing for the class 10 examination (CBSE, ICSE, & State boards). First, find the list of formulas from the electricity chapter. And, then find the solved numerical questions.

Formulas Used

1] In the following formulas, I is current, Q is the total charge, and t is the time duration through which the charge flows through the conductor.

I=Q/t

Q=I t

t = Q/I

2]
Work done = p.d. x charge moved
=> W = VQ
[here, p.d. = Potential difference]

3]
Q = n e
[n= total number of electrons, e = charge of an electron = 1.6 x 10-19 C]

4]
Heat energy H = I2Rt

5] Formulas for equivalent resistance:
Series circuit: Req=R1 + R2
Parallel circuit: 1/Req=1/R1 + 1/R2

6]
R=ρ l /A
Resistance = resistivity x length/cross-sectional area

Numerical Problems on Electricity chapter for Class 10

Question 1] If 20 C of charge pass a point in a circuit in 1 s, what current is flowing?

Solution
Given, Q = 20 C, t=1s
I=?
We know that: I=Q/t.
I=20/1 Amp =20 A.

Question 2] A current of 4 A flows around a circuit for 10 s. How much charge flows past a point in the circuit in this time?

Solution :
Given, I=4 A, C, t=10s
Q=?
We know that:
I=Q/t.
Q=It=4*10=40 C.

Question 3] What is the current in a circuit if the charge passing each point is 20 C in 40 s?

Solution :
Given, Q = 20 C, t=40 s
I=?
We know that:
I=Q/t.
Thus I=20/40=0.5 A.

Question 4] If a potential difference of 10 V causes a current of 2 A to flow for 1 minute, how much energy is transferred?

Solution :
Given: V = p.d. = 10 Volts,
I = 2 A,
t = 1 min = 60s.
We know that:
Q=I t.
Q=2×60.
Q=120 C. ………… (1)
Work done = Potential difference x charge moved.
=>W =VQ
W = 10 x 120 J
Work done = 1200 J.

Question 5]
(a) Write down the formula which relates electric charge, time and electric current.
(b) A radio set draws a current of 0.36 A for 15 minutes. Calculate the amount of electric charge that flows through the circuit.

Solution :
(a). Q =I t [I is the electric current, Q is the total electric charge, and t is the time duration through which the charge flows through the conductor]
(b).
Given data:
I=0.36 A,
t=15 min =900 seconds.

Q=I t
=0.36×900 C
= 324 C.

Question 6] If the charge on an electron is 1.6 x 10-19 coulombs, how many electrons should pass through a conductor in 1 second to constitute 1 ampere current?

Solution
Given: I = 1 A,
t=1 s
So, Q = I t= 1×1 C
Q= 1 C………………. (1)
Again, Q = n e [n= number of electrons to be found out, e = charge of an electron = 1.6 x 10-19 C]
=> n = Q/e = 1/[1.6 x 10-19 ]= 6.25×1018
So, the number of electrons =6.25×1018

Question 7] The p.d. across a lamp is 12 V. How many joules of electrical energy are changed into heat and light when :
(a) a charge of 1 C passes through it?
(b) a charge of 5 C passes through it?
(c) a current of 2 A flows through it for 10 s?

Solution:

For the given cases, V = Potential Difference = 12 Volts

a) Q = 1 C
W = VQ = 12 x 1 J = 12 J
Amount of electrical energy that changed into heat and light = 12 J
b) Q = 5 C
W = VQ = 12 x 5 J = 60 J
Amount of electrical energy that changed into heat and light = 60 J
c) I = 2 A, t = 10 s
Q= It = 2×10 C = 20 C
W = VQ = 12 x 20 = 240 J
Amount of electrical energy that changed into heat and light = 240 J

Question 8] In 10 s, a charge of 25 C leaves a battery, and 200 j of energy are delivered to an outside circuit as a result.
(a) What is the p.d. across the battery ?
(b) What current flows from the battery ?

Solution:
Given: t =10 s, Q = 25 C
Energy = 200 J
a) Energy = p.d x charge
pd = energy/charge = 200/25 V = 8 V
b) I = Q/t = 25/10 A = 2.5 A

Question 9] A flash of lightning carries 10 C of charge which flows for 0.01 s. What is the current? If the voltage is 10 MV, what is the energy?

Solution:
Given: Q = 10 C
t = 0.01 s
I = Q/t = 10 /0.01 A = 1000 A
The current is 1000 A [Answer]
Given p.d. = V = 10 MV
Energy = work done = W = VQ
So after putting values, W = VQ = 10 x 106 x 10 J = 100 x 106 J = 100 MJ

Question 10] An electric heater is connected to the 230 V mains supply. A current of 8 A flows through the heater.
(a) How much charge flows around the circuit each second ?
(b) How much energy is transferred to the heater each second ?

Solution:
V = 230 V
I = 8 A
a) t = 1 s
Q = I t = 8×1 C = 8 C
b) Energy = work done = W
W = VQ = 230 x 8 = 1840 J

Question 11] How many electrons are flowing per second past a point in a circuit in which there is a current of 5 amp ?

Solution
I = 5 A, t = 1 s
Q = It = 5×1 C = 5 C
Again, Q = ne
n = Q/e = 5/[1.6 x 10-19] = 31.25 x 1018

Question 12] A resistance of 20 ohms has a current of 2 amperes flowing in it. What potential difference is there between its ends ?

Solution :
R = 20 ohms
I = 2 A
We know that
V = IR
Thus,
V = 2 x 20 V
V = 40V

Question 13: A current of 5 amperes flows through a wire whose ends are at a potential difference of 3 volts. Calculate the resistance of the wire.

Solution :
I = 5 A
p.d., V = 3V
We know that V=IR
Thus,
3 = 5 x R
R = 3/5 = 0.6 ohm

Question 14: When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Calculate the value of the resistance of the resistor.

Solution :
V = 12 volt, I=2.5 x 10-3 A
We know that
V=IR
R=V/I
R=12/(2.5X10-3)
R=4.8 X 103ohm = 4800 ohm.

Question 15: What p.d. is needed to send a current of 6 A through an electrical appliance having a resistance of 40 Ω?

Solution:
I=6 A, R=40 ohm
We know that
V=IR
V = 6 x 40 = 240 V.

Question 16) A wire is 1.0 m long, 0.2 mm in diameter, and has a resistance of 10 Ω. Calculate the resistivity of its material.

Solution:
l = 1m
r = d/2 = 0.2/2 mm = 0.1 mm = 0.0001m,
R = 10 ohm
We know that,
R = ρ l /A
ρ = RA/l
=[10 x π x (0.0001)2]/1 = 31.4 x 10-8 Ωm

Question 17)
What will be the resistance of a metal wire of length 2 meters and area of cross-section 1.55 × 10-6 m2, if the resistivity of the metal is 2.8 × 10-8 Ωm?

Solution:
l=2m
A =1.55 × 10-6 m2
ρ = 2.8 × 10-8 Ωm
So, resistance R = ρ l /A = 2.8 × 10-8 x 2 /[1.55 × 10-6] Ω = 0.036 Ω

Question 18:
(a) State the relation between potential difference, work done and charge moved.
(b) Calculate the work done in moving a charge of 4 coulombs from a point at 220 volts to another point at 230 volts.

Solution :
(a)
Potential difference = Work done/Charge moved.
or
Work done= Potential difference x Charge moved
(b)
V1=220 V, V2=230V, Charge moved=Q=4C
Thus, the potential difference=V= V2 – V1 =230-220 =10 V.
We know that Work done = Potential difference x Charge moved
W=VQ
=>W= 10 x 4 J
Work done = 40 joules

Question 19:
(a) Name a device that helps to measure the potential difference across a conductor.
(b) How much energy is transferred by a 12 V power supply to each coulomb of charge which it moves around a circuit?

Solution :
(a) Voltmeter
(b) Given: Potential difference=V=12V, Charge moved=Q=1C
We know that,
Work done = Potential difference x charge moved
W=VQ
=>W= 12 x 1 = 12 joules
Since work done on each coulomb of charge is 12 joules, the energy given to each coulomb of charge is also 12 joules.

Question 20:
Three 2 V cells are connected in series and used as a battery in a circuit.
(a) What is the p.d. at the terminals of the battery
?
(b) How many joules of electrical energy does 1 C gain on passing through (i) one cell (ii) all three cells ?

Solution :
(a) If three cells of 2 volts each are connected in series to make a battery, then the total potential difference between the terminals of the battery will be 6V.
(b) (i) Given: p.d. of one cell = 2V, Charge moved = 1C
We know that
Work done = p.d. x charge moved = 2 x 1 J
Work done = 2 joules
(ii) Given: p.d. of 3 cells = 6V, Charge moved = 1C
Work done = p.d. x charge moved = 6 x1 J
Work done = 6 joule.

Question 21: If 3 resistances of 3 ohm each are connected in parallel, what will be their total resistance?

Solution:

R1= R2=R3= 3 Ω
Say, Equivalent resistance =R
As the resistances are in parallel, 1/R = 1/R1 + 1/R2 + 1/R3
=> 1/R = 1/3 + 1/3 + 1/3
=>1/R=1
=>R = 1 Ω

Question 22: What possible values of resultant resistance one can get by combining two resistances, one of value 2 ohm and the other 6 ohms?

Solution:
R1 = 2 Ω. R2 = 6 Ω
Case 1: (Parallel combination)
1/R = 1/R1 + 1/R2
=>1/R = 1/2 + 1/6
=>1/R = 4/6
=>R = 6/4=1.5 Ω
Case 2: (Series combination)
R=R1 + R2
=>R = 2+ 6 = 8 Ω

Question 23: Show how you would connect two 4-ohm resistors to produce a combined resistance of
(a) 2 ohms
(b) 8 ohms.

Solution :
(a) By connecting in parallel: Since equivalent resistance will be
1/ R = 1/4 + 1/4 = 2/4 = 1/2
Therefore, R = 2 ohm
(b) By connecting in series : Since equivalent resistance will be
R = 4 ohm + 4 ohm = 8 ohm.

Question 24: A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor?

Solution:

Question 25: Two resistors, with resistances 5 Ω and 10 Ω respectively are to be connected to a battery of emf 6 V so as to obtain :
(i) minimum current flowing (ii) maximum current flowing
(a) How will you connect the resistances in each case?
(b) Calculate the strength of the total current in the circuit in the two cases.

Solution :
Given: Two resistors with resistances R1=5 ohm and R2=10 ohm, V=6 volt
(a) For minimum current these two should be connected in series. For maximum current, these two should be connected in parallel.

(b)In series,
Total resistance = 5+10 = 15 ohms
Therefore in series total current drawn = V/R = 6/15 = 0.4 amps

In parallel,
Total resistance R is given as
1/R=1/R1+1/R2
1/R=1/5+1/10
1/R=3/10
R=10/3 ohm
Therefore total current drawn by the circuit = V/R = 6/(10/3) =1.8 amps.

Question 26: A resistor has a resistance of 176 ohms. How many of these resistors should be connected in parallel so that their combination draws a current of 5 amperes from a 220-volt supply line?

Solution:

I = 5 A
V = 220 V
Hence, resistance R = V/I = 220/5 Ω = 44 Ω

Resistance of available resistor = 176 Ω
But, resistance required is 44 Ω, which is less than 176 Ω.
So to get the required resistance, available resistors are to be connected in parallel.
Let, the required number of resistors is n.
In parallel connection, Req =R/n
=>44=176/n
=>n = 176/44 = 4
So 4 resistors should be connected in parallel. [Answer]

Question 27: A p.d. of 4 V is applied to two resistors of 6 Ω and 2 Ω connected in series. Calculate :
(a) the combined resistance
(b) the current flowing
(c) the p.d. across the 6 Ω resistor

Solution:

Given: V = 4V,
R1 = 6 Ω, R2 = 2 Ω [connected in series]
(a) Combined resistance R = R1 + R2 = 6 + 2 = 8 Ω
(b) Current I = V/R = 4/8 = 0.5 A
(c) PD across 6 Ω resistor = I R1= 0.5 x 6 V = 3 V

Question 28: A p.d. of 6 V is applied to two resistors of 3 Ω and 6 Ω connected in parallel.
Calculate:
(a) the combined resistance
(b) the current flowing in the main circuit
(c) the current flowing in the 3 Ω resistor.

Solution:

Question 29: A 4 Ω coil and a 2 Ω coil are connected in parallel. What is their combined resistance? A total current of 3 A passes through the coils. What current passes through the 2 Ω coil?

Solution:

Question 30: Two resistances when connected in parallel give the resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

Solution:

Question 31: A bulb is rated as 250 V; 0.4 A. Find its: (i) power, and (ii) resistance.

Question 32: ] For a heater rated at 4 kW and 220 V, calculate :
(a) the current,
(b) the resistance of the heater,
(c) the energy consumed in 2 hours, and
(d) the cost if 1 kWh is priced at ₹ 60.

Solution:

Question 33) Which uses more energy: a 250 W TV set in 1 hour or a 1200 W toaster in 10 minutes?

Solution:

Question 34) A resistance of 25 Ω is connected to a 12 V battery. Calculate the heat energy in joules generated per minute.

Solution
Given: R = 25 Ω, V = 12 V, time t = 1 min = 60 s
To find out: Heat energy H =?

Current I = V/R = 12/25 A = 0.48 A
As, H = I2Rt
H= (0.48)2 x 25 x 60 J = 345.6 J

Question 35) 100 joules of heat is produced per second in a 4-ohm resistor. What is the potential difference across the resistor?

Solution
Given, H = 100 J, t = 1 s, R = 4 Ω
H = I2Rt
=> I2 = H/[Rt] = 100/[4x 1] = 25
I = 5 ……….. (1)
Say, potential difference across the resistor = V
=>V= IR = 5 x 4 V= 20 V

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