1. A voltmeter has high resistance and it is connected in parallel True or false?
2. What is resistance of the conductor?
3. What is the SI unit of resistivity?
4. Wire of length 3 m and area of cross -section 1.7X10-6 m
2
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
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
5. Arrange the samples in increasing order of conductivity
a) which of these is best conductor?
b) Which are these is best insulator?
6. How does the resistivity of alloys compare with those of pure metals form which they may have
formed?
7. Define 1 volt. Express it in terms of SI unit of work and charge. Calculate the amount of ener gy
consumed in carrying a charge of 1 coulomb through a battery of 3 volts.
Answers
Answer:
1. It is true that a voltmeter has high resistance and it is connected in parallel.
This is because if the voltmeter has low resistance, it will act like a barrier and will draw current from the source in the circuit.
2. Resistance of a conductor can be defined as the difference of the potential across the two ends of the conductor required to pass a unit electric current through it.
3. The SI unit of resistivity is ohm⋅metre (Ω⋅m) .
4. a) The formula for resistivity of the wire = (R × A) ÷ l
where R = electrical resistance of the wire.
A = cross sectional area of the wire.
l = length of the wire.
b) The resistivity of the wire = (R × A) ÷ l
= (3 × 10⁻² × 1.7 × 10⁻⁶) ÷ 3
= 1.7 × 10⁻⁸ ohm.
5. The samples in increasing order of conductivity are : Y<X<Z.
a) Z is the best conductor among these.
b) Y is the best insulator among these.
6. The resistivity of alloys is higher than the pure metals. When several metals are combined, the atomic arrangement changes and the number of free electrons which produces more barrier to their flow.
7. 1 volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points.
1 volt can also be expressed as 1 joule of work done to carry 1 coulomb of charge between two points.
We know that,
V = W ÷ Q
or W = V × Q
where V = 3 volts.
Q = 1 coulomb.
Therefore,
W = 3 × 1
= 3 joule