Class 10 Science Worksheet for Electricity

We know, $P = VI$
And $I= \frac {V}{R}$
So
$P= \frac {V^2}{R}$
$R = \frac {V^2}{P}$

Case Maximum Heating
At maximum heating, power is P = 840 W
Working voltage is 220 V
So, resistance at maximum heating is = (220)²/840 = 57.6 Ohm
Now I = P/V
So, current at maximum heating is = 840/220 = 3.8 A
Case Minimum Heating
At minimum heating, power is = 360 W
The resistance at minimum heating will be = (220)²/360 = 134.4 Ohm
The current at minimum heating will be = 360/220 = 1.6 A

Copper and aluminium wires are usually employed for electricity transmission for the below reasons

(a)These are extremely good conductors
(b) They have a low value of resistivity.So while transmission over long distances, electricity is not dissipated in form of heat and energy is, hence, not wasted.
(c)these are ductile and can be drawn in the form of fine wire
(d) They are less expensive than the best conductors, silver and gold.
(e)They are extremely flexible and do not degrade under electrical transmission as do other metals.

(i) Let t be the time during which charge Q flows. Now when charge Q moves against the potential difference V , then the amount of work is given by


Therefore the source must supply energy equal to VQ in time t. Hence power input to the circuit by the source is

The energy supplied to the circuit by the source in time t is P×t that is, VIt.
This is the amount of energy dissipated in the resistor as heat energy.
Thus for a steady current I flowing in the circuit for time t , the heat produced is given by

Applying Ohm's law to above equation we get

This is known as Joule's Law of heating
ii) Now H=V²t/R
Here H=220J t=1 sec R=8 ohm
V²=1760
V=42 Volt

(a) When electric current passes through a high resistance wire, the wire becomes and produces heat. This is called heating effect of current. electric Iron,Electric Heater are two appliances which work on this effect
(b) P=VI
=220X.50=110W
(c) Usage per day= 400 x 8 = 3200 Whr =3.2 KWh
(d) Kilowatt is measure of Power while Kilowatt hour is the measure of energy

Work done = Potential difference charge
Here P.D=6V and Q=1C
So work done=6J

(a) Resistance of the wire is given by
$R=\rho \frac{l}{A}$
Here R=12 ohm,resistivity =1.6 X 10^-8ohm m. And $A =\pi (d/2)^2=\pi (.0005/2)^2$
So
$l= \frac{RA}{\rho}$
Substituting the above values
l=147.18 m
(b) Now when diameter is halved ,length remains same
$A =\pi (d/2)^2=\pi (.00025/2)^2$ and l=147.18 m
$R=\rho \frac{l}{A}$
Substituting all the values we get
R=48 ohm

Let the resistivity of the material is ?, length of the wire initially l and the area of cross section is A. Then the resistance of the wire will be,
$R=\rho \frac{l}{A}=4$


Now cut the wire in equal four pieces and stretch it to the initial length l. When we stretch it the new length will be same of each piece but the area get reduced to one fourth ,So resistance of that piece would become
$R_1=\rho \frac{l}{A/4}=4\times 4=16$

Now, connect these four pieces in parallel. Then the equivalent resistance will be,
$\frac {1}{R_equiv}=\frac {1}{R_1} +\frac {1}{R_2}+ \frac {1}{R_3}+\frac {1}{R_4}$
$\frac {1}{R_equiv}=\frac {1}{16} +\frac {1}{16}+ \frac {1}{16}+\frac {1}{16}$
R_equiv=4 ohm

$P = V \times I = 2 \times 500 =1000W$
$E= P \times t= 1000 \times 300 = 3 \times 10^5$ J

$P=VI$
$= 220 \times 10$
= 2200W

Electricity consumption in a day
$E=P \times T$
$=2200 \times 5$#
= 11000W=11 KWH /day
Usage per month=11*30 = 330KWH
Cost of the Electricity= 330* 2.50= RS.825

I=.2 A
V=220 V
Now
t= 1 hour = 3600 sec
Now
$I = \frac {q}{t}$
or
$q = I \times t = .2 \times 3600 = 720 C$

$R = \frac {V^2}{P} = 484 \omega$
Now
$P = \frac {V^2}{R} = 25W$

Given R₁ = 10 ohm, R₂ = 20 ohm, and R₃ = 30 ohm $R=R_1 + R_2 + R_3 = 10 +20 + 30 = 60 /omega$ Now from Ohm's Law $V=IR$ $I = \frac {V}{R} = .1 A$

I=.2 A
Now
t= 5 hour = 18000 sec
Now
$I = \frac {q}{t}$
or
$q = I \times t = .2 \times 18000 = 3600 C$

1 Ampere means 1 C/sec.So we need to find the number of electrons in 1 C
Now Charge on 1 electron=$1.6 \times 10^{-19} C$

So Number of electrons on 1 C
$= \frac {1}{1.6 \times 10^{-19}} =6.25 \times 10^{18}$

R=5 ohm, V= 6 Volt, I=?
$I = \frac {V}{R} = 1.2 A$
Now Energy dissipated
H= VIt= 300 J

Power is given by
$P = \frac {V^2}{R}$
So, resistance can be obtained as
$R = \frac {V^2}{P}$
For Electric Lamp A
$R = \frac {V^2}{P} =\frac {220^2}{40} =1210 \omega $
For Electric Lamp B
$R = \frac {V^2}{P} =\frac {220^2}{60} =806.66 \omega $
So, Lamp A has higher Resistance
Now Lamp B will glow brighter when connected with 220 V as the power is higher than lamp A

$H=I^2Rt$
=5^2 \times 12 \times 120 =3.6 \times 10^4 J$

$P=VI$
or $I= \frac {P}{V} = 9.09 A$
Thus, the electric geyser draws current much more than the current rating of 5 A
So, we cannot run this electric geyser on the 5 A line

$P=\frac {V^2}{R}$
$R=\frac {V^2}{P}$
For one bulb, resistance would be
$R=\frac {220^2}{100}=484 \omega$

Current will be given as
I=V/R = 220/484 =.45 A

Now One bulb drawing the 0.45 amps
11 bulbs drawing the 4.95 amps (<5 amps)
So, the number of 100 watts bulbs to connect with out blowing the fuse are 11.

$P =VI$
or $I = \frac {P}{V}$
Fore bulb A
I=.75 A
Fore bulb B
I=.5 A
Higher Power rating bulb will consume more energy, So Bulb A

$P =VI$

or $I = \frac {P}{V}$

I= 2000/220 = 9.09 A
Thus,10 A fuse should be used with electric Iron

$P =VI$

or $I = \frac {P}{V}$
I= 4000/220 = 18.18 A
$P =\frac {V^2}{I}$

or $R = \frac {V^2}{P}=12.1 \omega$
Now Energy consumed in 1 hour
= P X time = 4KWH

Battery -> It is used to supply electric current to the circuit
Ammeter -> It is used to measure electric current
Voltmeter -> It is used to measure Potential difference in the circuit

Electric resistance of a conductor is the obstruction offered by the conductor to the flow of the current through it.
Electric resistance depends on lenght,area,resistivity and temperature.It does not depend on current. Infact current is determined by the voltage and resistance. So Resistance would be same in both the cases

Ammeter is connected in series while voltmeter is connected in Parallel. If the are interchanged,incorrect reading will be specifoed

Potential difference accross 1 Ω resistor = IR=1 V
Same potential difference will be there in other resistors also
So Current in 2 Ω resistor= 1/2 =.5 A
Similarly current in 3 Ω resistir= 1/3=.333 A

$R=\rho \frac {L}{A}$
or
$L = \frac {RA}{\rho}$
$= \frac { 10 \times 3.14 \times 10^{-8}}{50 \times 10^{-8}}$
$=.628 $m

When in Parallel , equivalent reistance
$ \frac {1}{R_eq}=\frac {1}{R} + \frac {1}{R} + \frac {1}{R}$
R_eq = R/3

When in Series, equivalent resistance
R_eq = 3R,

So when connected in series, maximum resistance is obtained and when connected in parallel ,minimum resistance is obtained

$\frac {1}{R_eq}= \frac {1}{2} + \frac {1}{3} + \frac {1}{6} $
R_eq= 1Ω

a. This flow of charge in metallic wire due to the potential difference between two conductors used is called electric current. It SI unit is Ampere
1 Ampere = 1 columb /1 sec
b. Conventional current will flow from A to B as it is taken as the flow of positive charge
c. We know 1 ampere meams 1 coulomb per sec. So we need to calculate the number of electrons in 1 coulomb
$n= \frac {1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} $