Questions - Answers
Very Short Answer type Questions
1. Define mean free path of electron in a conductor.
A: Mean free path of electron: The average distance covered by an electron between two successive collisions is called the mean free path.
2. State Ohm's law and write its mathematical form.
A: Ohm's law: At constant temperature, the electric current flows through a conductor is proportional to the potential difference across its ends.
V = iR
3. Define resistivity or specific resistance.
A: Resistivity or Specific Resistance: Resistance of a conductor of unit length (1 m) and of unit area of cross section (1 m2) is called resistivity (ρ) of material of the conductor.
Its SI unit is Ω m.
4. Define temperature co - efficient of resistance.
A: Temperature co - efficient of resistance: The ratio between the change of resistance of a conductor per unit rise of temperature and its initial resistance is called temperature co - efficient of resistance (α) of the conductor.
RT = R0 [1 + α (T - T0)]
The SI unit of temperature co - efficient of resistance is K-1.
5. Why is manganin used for making standard resistors?
A: For manganin, temperature co - efficient of resistance (α) is very very less. So change in resistance of the manganin wire with temperature is almost negligible. So it is used to prepare standard resistance.
6. Write the colour code of a carbon resistor of resistance 23 Kilo Ohms.
A: Red stands for 2, Orange stands for 3.
23 Kilo Ohms = 23 × 103 Ohms
7. Why are household appliances connected in parallel?
A: Advantages of parallel connection:
1) Different appliances can take different currents depending on their resistances.
2) If one of the appliances fails or burns out, other appliances work as usual. If the appliances are connected in series, same current flows through all of them.
If one of them burns out, others won't work. That is why, household appliances are connected in parallel only.
8. Derive an expression for the effective resistance when three resistors are connected in (i) series ii) parallel.
A: Resistors in Series:
When resistors are connected end to end such that same current passes through them, they are said to be connected in series. Let us suppose that three resistors of resistances R1, R2 and R3 are connected in series as shown in the diagram. Same current I passes through them.
From Ohm's Law:
Potential drop across R1 is V1 = IR1
Potential drop across R2 is V2 = IR2
Potential drop across R3 is V3 = IR3
Total potential drop across the combination is V = V1 + V2 + V3 or V = IR1 + IR2 + IR3
If the combination acts as a resistor of equivalent resistance R with current I.
V = IR
IR = IR1 + IR2 + IR3
or R = R1 + R2 + R3
Resistors in parallel: 
When resistors are connected end to end such that same potential drop is formed on them, they are said to be connected in parallel.
Let us suppose that three resistors of resistances R1, R2 and R3 are connected in parallel as shown in the diagram. Some potential drop V exists on each of them.
From Ohm's law:
9. State Kirchoff's laws for an electrical network. Using these laws deduce the condition for balance in a Wheatstone bridge.
A: Kirchoff's Rules:
Junction Rule: At any junction in an electric circuit, the sum of currents entering the junction is equal to the sum of currents leaving the junction.
Loop Rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.
Wheatstone's Bridge:
The circuit shown in the figure is called Wheatstone's bridge. It has four resistors R1, R2, R3 and R4. AC is battery arm, BD is galvanometer arm. The galvanometer G detects current.
If the resistors are adjusted such that the galvanometer current Ig = 0, the bridge is said to be balanced.
Applying Kirchoff's junction rule at D, we get
I1 = I3 ...................... (1)
Applying Kirchoff's junction rule at B, we get
I2 = I4 ...................... (2)
Applying Kirchoff's loop rule to closed loop ADBA, we get
- I1 R + 0 + I2R2 = 0
Applying Kirchoff's loop rule to closed loop CBDC, we get
I4R4 + 0 - I3R3 = 0
But I1 = I3 & I2 = I4
I2R4 - I1R3 = 0
This is the balance condition of Wheatstone's bridge to make Ig = 0
10. Show the variation of current versus voltage graph for (GaAs) and mark the (i) Non - linear region (ii) Negative resistance region.
A: The current voltage graph of a semi conductor like Gallium - Arsenide (GaAs) is as shown in the figure. When voltage increases, current in it varies peculiarly.
i) In lower part of the voltage change current increases but not propartionally. This is called non - linear region.
ii) In later part if voltage increases, current decreases. This is called negative resistance region.
Long Answer type Questions
1. State the working principle of potentiometer. Explain with the help of circuit diagram how the e.m.f. of two primary cells are compared by using the potentiometer.
A: Potentiometer: Potentiometer is a device used to measure emf of a cell without drawing any current from it.
Construction: Potentiometer consist of a wooden board on a uniform manganin wire of length 4 m is fixed in parallel rows between two binding screws A and C by the side of a meter scale. A jockey is also provided with it to make a contact at any point of the wire.
Principle: The potential drop between the starting point and any point of the wire is directly proportional to the length l of the wire between the two points.
ε = φl
Where φ is potential drop per unit length of the potentiometer wire.
Comparision of emfs of two cells: The primary circuit consists of a battery B, a rheostat Rh, a plug key K1. Rh can be adjusted to provide a constant current to the potentiometer wire. The secondary circuit of the consist of two way key, two cells (ε1, ε2) and a galvanometer G as shown in the diagram.
First the key K1 is closed to give a current to potentiometer wire. The cell ε1 is included in the circuit by connecting 1 and 3 of the two way key.
ε = φ l
Where φ is potential drop per unit length of the potentiometer wire.
Potentiometer experiment to determine internal Resistance of a Cell: The primary circuit provides constant current (I) to the potentiometer wire. The secondary circuit of the potentiometer is as shown in the diagram.
First the key K1 is closed. Keeping K2 open the position of the jockey is adjusted until the galvanometer shows null deflection. Its balancing length l1 of the wire is noted. Now the key K2 is closed to include R in the circuit.
Again the position of the jockey is adjusted for null deflection of the galvanometer. Balancing length of the wire l2 is noted.
In the first case, the cell is in the circuit,
ε = φl1 .................... (1)
Where φ is potential drop per unit length of the potentiometer wire.
The balancing length l1 of the wire is found by adjusting position of its jockey for null deflection of the galvanometer G.
ε1 = φl1 .................... (1)
Now 2, 3 are connected to do the experiment with cell ε2. Its balancing length l2 is noted.
ε2 = φl2 .................... (2)
Dividing Eq (1) by Eq (2), we get
The emfs of the given cells can be compared by using this equation.
2. State the working principle of potentiometer. Explain with the help of circuit diagram how the potentiometer is used to determine the internal resistance of the given primary cell.
A: Potentiometer: Potentiometer is a device used to measure emf of a cell without drawing any current from it.
Construction: Potentiometer consists of a wooden board on which a uniform manganin wire of length 4 m is fixed in parallel rows between two binding screws A and C by the side of a meter scale. A jockey is also provided with it to make a contact at any point of the wire. The potential drop between the starting point and any point of the principle: wire is directly proportional to the length l of the wire between two points.
ε = φl1 .................... (1)
In the second case, R is in parallel to the cell. The terminal voltage is given by
V = φl2 .................... (2)
Dividing equation (1) by equation (2)
Internal resistance r can be calculated from this formula.
3. In a house three bulbs of 100 W each are lighted for 4 hours daily and six tube lights of 20 W each are lighted for 5 hours daily and a refrigerator of 400 W has worked for 10 hours daily for a month of 30 days. Calculate the electricity bill if the cost of one unit is Rs. 4.00
A: Work in KWH = Power in KW × Time in hours
W = 30[1.2 + 0.6 + 4] KWH
W = 30 × 5.8 KWH
W = 174 KWH
Electricity Bill = W in KWH × cost per unit = 174 × 4.00 = Rs. 696
