# Electricity

## Exemplar Problem

### Multiple Choice Questions 1

Question 1. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams in following figure. The current recorded in the ammeter will be

1. maximum in (i)
2. maximum in (ii)
3. maximum in (iii)
4. the same in all the cases

Answer: (d) the same in all the cases

Question 2. In the following circuits, heat produced in the resistor or combination of resistors connected to a 12 V battery will be

1. same in all the cases
2. minimum in case (i)
3. maximum in case (ii)
4. maximum in case (iii)

Answer: (c) Maximum in case (ii)

Explanation: Heat produced is directly proportional to the resistance. In case two the total effective resistance is equal to 4Ω which is maximum in all the given three conditions. Thus, in this case heat produced would be maximum.

Question 3. Electrical resistivity of a given metallic wire depends upon

1. its length
2. its thickness
3. its shape
4. nature of the material

Answer: (d) nature of the material.

Question 4. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly

1. 1020
2. 1016
3. 1018
4. 1023

Explanation: Electric current, I = 1A,
Time, t = 16 second
Charge, Q =?

I= Q/t

⇒1A= Q/(16s)

⇒Therefore, Q=16 coulomb

Since, 1.6xx10^(-19) coulomb of charge contains 1 electron

Thus,1 C of charge have number of electrons equal = 1/(1.6xx10^(-19))

Therefore,16 coulomb of charge contains 1/(1.6xx10^(-19))xx16 electrons

=(1xx16)/(16xx10^(-20)) electrons = 1020 electrons

Question 5. Identify the circuit in which the electrical components have been properly connected.

1. (i)
2. (ii)
3. (iii)
4. (iv)

Question 6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

1. 1/5 Ω
2. 10 Ω
3. 5 Ω
4. 1 Ω

Explanation: When resistors are connected in series, the effective resistance becomes the sum resistance of all resistors which is maximum.

The total effective resistance when connected in series = 1/5 Ω+1/5 Ω+1/5 Ω+1/5 Ω+1/5 Ω

=(1+1+1+1+1)/5 Ω= 5/5=1Ω

Question 7. What is the minimum resistance which can be made using five resistors each of 1/5 Ω?

1. 1/5 Ω
2. 1/(25) Ω
3. 1/(10) Ω
4. 25 Ω

Answer: (b) 1/(25)Ω

Explanation: When resistors are connected in parallel, the total effective resistance is reduced and that comes minimum.

Let total effective resistance = R

Therefore,1/R=5/1 Ω+5/1 Ω+5/1 Ω+5/1 Ω+5/1 Ω= (5+5+5+5+5)/1 Ω=25/1 Ω=25Ω

Thus,R= 1/(25) Ω

Question 8. The proper representation of series combination of cells (given figure) obtaining maximum potential is

1. (i)
2. (ii)
3. (iii)
4. (iv)