#### Extra Questions for Class 10th: Ch 12 Electricity (Science) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 1 Mark

Q1. What will happen if switch of a particular bulb among three bulbs connected in series is switched off?
No current will flow

Q2. How voltmeter should be connected to a resistor in order to find potential difference across it.?
It is connected parellel to resistor.

Q3. Why resistivity is also known as specific resistance?
Resistivity is the resistance of a material at specific condition, unit length and unit cross section.

Q4. On what principle is an electric bulb based?
Heating effect of current.

Q5. Why are coils of electric toasters and electric irons made of alloy rather than pure metals?
Alloys are preferred to make coils of toasters and irons because they have higher resistivity and do not oxidise easily at high temperature.

Short Answer Questions-I (SAQs-I): 2 Marks

Q1. An ammeter is always connected in series across a circuit element. What happens when it is connected in parallel with a circuit element?

An ammeter is a low-resistance device. When it is connected in parallel, the resistance of the circuit reduces considerably. Therefore, a large current flows through the circuit, by virtue of circuit element may damage.

Q.2 Which is analogous to temperature : a current or electric potential?

Electric potential is analogous to temperature. As heat flows from higher temperature to lower temperature, so charges flow from higher potential to lower potential.

Q3. Out of the two wires given, which one is having higher resistance? Given both wire is made up of same material and temperature is constant? Both are having circular cross section whose radius is given. Since, both wires are having same material, their resistivity is same at same temperature. It is given as ρ(Let). Q4. 320J of heat is produced in 10s in a 2Ω resistor. Find amount of current flowing through the resistor?

Given,
Heat produced H=320J
Time taken t=10s
Resistance of resistor R=2Ω
Current I (let)
Since, H=I2 Rt
I2=H/Rt=320J/(2×10s)=(16A)2
I=4A
So, 4A of current is flowing through the resistor.

Q5. Write the factors, the resistance of a wire depends upon.

The factors are
(i)Temperature: Resistance increases with increase in temperature and vice versa.
(ii) Length of conductor: Resistance increase with increase in temperature and vice versa.
(iii)Cross sectional Area: Resistance decreases with increase in cross sectional Area and vice versa.

Short Answer Questions-I (SAQs-II): 3 Marks

Q10. The resistance of a wire of 0.01 cm radius is 10Ω. If the resistivity of the material of the wire is 50 x108 ohm meter, find the length of the wire. Q 1. Derive an expression for electric energy consumed in a device in terms of I,t, and R, where I is the , I is the current drawn in resistance R and t is the time for which the current flows ?

We know that work done = W = QV
V is potential difference, Q is charge passed.
V=IR
Now, we know current I=Q/t
Q=It
In t times Q charge has been passed.
W=It.IR
W=I2tR
So, electric energy consumed is given as
W=I2Rt

Q 2. Give reason for the following : Why are copper and aluminum wires used as connecting wires ? Why is tungsten used for filament of electric lamps? Why is lead-fin alloy used for fuse wires ?

These are good conductors of resistance low resistivity. Also, they are much cheaper than silver which is having highest conductivity.
They are having very high melting point and high resistivity.
It is having required melting point, so that it melts at required temperature.

Q 3. State the different ways, three resistors each having resistance r can be connected so that they will draw maximum current from the circuit. Explain

The three resistors will draw maximum current when are equivalent resistance will be minimum. This is the case when the resistors are connected in parallel with each other. In this case, the equivalent resistance will be R
Then,
1/R=1/r+1/r+1/r
1/R=3/r
R=r/3
In this case, maximum current will be drawn from the battery.
The minimum current will be drawn from the circuit, when resistance is maximum. This is the case, when resistances are connected in series.
The equivalent resistance of the system will be R’=r+r+r=3r
In this case, minimum current will be drawn from the battery.

Q 4. A metallic wire of resistance R is cut into ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. What will be the effective resistance of the combination?

The resistance of a conductor is directly proportional to the length of the conductor. The resistance of each piece of wire, when it is cut into ten parts of equal length, r=R/10, as we know resistance is directly proportional to the length of wire.
Two such pieces are when joined in series, the equivalent resistance of two part will be r+r=R/10+R/10=R/5
5 such elements are connected in parallel. Therefore, total resistance R’ will be equivalent resistance of all resisitors.
1/R' =5/R+5/R+5/R+5/R+5/R=5×5/R=25/R
R'=R/25

Long Answer Questions (LAQs): 5 Marks

Q1. When we say the resistors are in parallel? Write the expression for the resistors in parallel. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

When we can say that resistors are in parallel if same potential difference is applied across them.
For three resistors, if we connect them in parallel, the equivalent resistance is R
1/R=1/R1+1/R2+1/R3
For x number of resistors of resistance 176Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm's law as V=IR
R = V/I
where,
supply voltage, V = 220 V
Current, I = 5 A
Equivalent resistance of the combination = R,
Given as, Q2. Two wires X and Y are of equal length and have equal resistances. If the resistivity of X is more than that of Y, which wire is thicker and why ? For the electric circuit given below calculate Current in each resistor
Total current drawn from battery
Let length of each resistor be l, let resistivity of wire X be  ρx, and resistivity of wire Y be ρy.
Let cross sectional Area of X be Aand of Y is  AY.
ρxy
But resistance of each wire X and Y be
Rx=Ry
ρx l/Ax=ρy l/Ay.
ρx.Ayy.Ax
Since,
ρxy
Therefore, Ax>Ay
Hence, the cross sectional area of wire X is more than cross sectional area of wire Y. Since, all three resistors are in parallel, current through each resistor, is potential difference applied, divided by the resistance of each resistor.
Current through 3Ω
6V/3Ω=2A
Current through 10Ω
6V/10Ω=0.6A
Current through 5Ω
6V/5Ω=1.2A
Total current is current drawn by the circuit is the sum of all three currents
2A+0.6A+1.2A=3.8A