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Class 10 Science Numerical Questions for Electricity





Given below are the Class 10 Science Numerical Questions for electricity
a) Concepts questions
b) Calculation problems
c) Multiple choice questions
d) Long answer questions
e) Fill in the blank's
 1)  A wire of length 3 m and area of cross-section 1.7X10-6 m2 has a resistance 3X10-2 ohm.
a) What is the formula for resistivity of the wire and what is the unit of it
b) Calculate the resistivity of the wire
Ans)
a) Resistivity of the wire is given by
$\rho =R\frac{A}{L}$
And It unit is Ohm-m
b) In this case
L=3 m
A=1.7X10-6 m2
R=3X10-2 ohm
So
ρ=1.7X10-8 Ohm-m
2) The table given below shows the resistivity of three Material X, Y and Z?
Samples
X
Y
Z
Resistivity
3 X10-9
11.1 X10-6
18X10-17
a) Arrange the samples in increasing order of conductivity
b) Which of these is best conductor?
c) Which are these is best insulator?
Ans)
a) Conductivity is inversely proportional to resistivity
So
Y< X< Z
b) Z is the best conductor as it has least resistivity
c) Y is the best insulator as it has highest resistivity
3)  There are m resistor each of resistance R. First they all are connected in series and equivalent resistance is X. Now they are connected in parallel and equivalent resistance is Y.  What is the ratio of X and Y?
Ans)
Series combination
X=R+R+R+…….= mR
Parallel combination
$\frac{1}{Y}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+...=\frac{m}{R}$
Or  Y=R/m
So X : Y=m2 : 1
4) We have  four resistors A ,B ,C and D of resistance  4ohm,8ohm ,12 ohm and 24 ohm respectively?
1
Lowest resistance which can be obtained by combining these four resistors
2
highest resistance which can be obtained by combining these four resistors
Ans)
Lowest resistance is obtained in parallel combination
$\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}$
Or R=2Ω
Highest resistance is obtained in series combination
R=4+8+12+24=64Ω
5)  Three resistors  5Ω, 10Ω and 30Ω are connected in parallel with the battery of Voltage  6V?
S.no
Questions
1
The value of current across each resistor
2
The value of Potential difference across each resistor
3
Total current in the circuit
4
Effective resistance of the circuit
Ans)
Potential difference remains same across parallel combination
So current in each resistor is calculated as
I1=V/R1=6/5=1.2 A
I2=V/R2=6/10=.6 A
I3=V/R3=6/30=.2 A
Total current in the circuit
I=I1+I2+I3=1.2+.6+.2=2 A
Effective resistance
$\frac{1}{R}=\frac{1}{5}+\frac{1}{10}+\frac{1}{30}$
  • R= 3Ω
6) An electric bulb draws a current of .8 A and works on 250 V on the average 8 hours a day.
a) Find the power consumed by the bulb
b) If the electric distribution company changes Rs 5 for 6 KWH, what is the monthly bill for 60 days
Ans)  Power of the electrical bulb is given by
P=V X I=.8 A X 250 V= 200 W =.2 KW
Total energy consumption by the bulb in 60 days
E=P X t= .2 X8 X60=96 KWH
So cost will be =5X96/6=80 Rs
7)

A1 ,A2 ,A3 and  A are ammeters connected in the circuit
B1 ,B2 and B3 are three identical bulbs
They all are connected to Voltage source as shown in Figure
When the three bulb are working good and glowing ,the current recorded in Ammeter A is 6 A
Answer Following questions
a) Same amount of current will go through each Bulb.  And the value is 2 A .True or False
b) If the Bulb B3 ­is blown away, the bulb B­­1 and B2 will start glowing more. True or False
c) What will happen to all the ammeter reading if Bulb B1 is blown away
d) The current shown in Ammeter A remains  even any bulb goes down. True or False
Ans)
a) Since Bulb are identical and connected in parallel with Voltage. Same current will flow through each bulb. Since the total current is 6 A.Individual current will be 2 A
b) If the Bulb B is blown away, The potential difference across other bulb still remains same, So same current will flow and they will glow as it is. No change
c) when Bulb B­ goes down, the current in that part become zero.
So reading of Ammeter A1 becomes zero
Reading of Ammeter A2  will remain same i.e. 2 A
Reading of Ammeter A3  will remain same i.e. 2 A
Reading of Ammeter A will be = 2+2=4 A
d) As shown above, the reading of Ammeter A will change
8)  Gave the formula for each
1
Ohm’s Law
2
Resistance in terms of Length,Area,resistivity
3
Current in terms of Resistance and Voltage
4
Equivalent Resistance for Resistors in Series
5
Equivalent Resistance for Resistors in Parallel
6
Power produced in the resistance
Solutions
1
Ohm’s Law
V=IR
2
Resistance in terms of Length,Area,resistivity
$R=\rho \frac{L}{A}$
3
Current in terms of Resistance and Voltage
$I=\frac{V}{R}$
4
Equivalent Resistance for Resistors in Series
R=R1+R2+R3…..
5
Equivalent Resistance for Resistors in Parallel
$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$
6
Power produced in the resistance
P=I2R

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