Question 1

Question 1: You are given four bulbs of 25 W, 40 W, 60 W, and 100 W of power, all operating at 230 V. Which of them has the lowest resistance? (A) 25 W (B) 40 W (C) 60 W (D) 100 W

Accepted Answer

(D) 100 W

Question 2

Question 2: Which of the following is an ohmic conductor? (A) transistor (B) vacuum tube (C) electrolyte (D) nichrome wire

Accepted Answer

(D) nichrome wire

Question 3

Question 3: A rheostat is used (A) to bring on a known change of resistance in the circuit to alter the current. (B) to continuously change the resistance in any arbitrary manner and there by

Accepted Answer

(B) to continuously change the resistance in any arbitrary manner and there by alter the current.

Question 4

Question 4: The wire of length L and resistance R is stretched so that its radius of cross-section is halved. What is its new resistance? (A) 5R (B) 8R (C) 4R (D) 16R

Accepted Answer

(D) 16R

Question 5

Question 5: Masses of three pieces of wires made of the same metal are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratios of their resistances are (A) 1 : 3 : 5 (B

Accepted Answer

(D) 125 : 15 : 1

Question 6

Question 6: The internal resistance of a cell of emf 2 V is 0.1 Ω, it is connected to a resistance of 0.9 Ω. The voltage across the cell will be (A) 0.5 V (B) 1.8 V (C) 1.95 V (D) 3V

Accepted Answer

(B) 1.8 V

Question 7

Question 7: 100 cells each of emf 5 V and internal resistance 1 Ω are to be arranged so as to produce maximum current in a 25 Ω resistance. Each row contains equal number of cells. The number

Accepted Answer

(A) 2

Question 8

Question 8: Five dry cells each of voltage 1.5 V are connected as shown in diagram What is the overall voltage with this arrangement? (A) 0 V (B) 4.5 V (C) 6.0 V (D) 7.5 V

Accepted Answer

(B) 4.5 V

Question 9

Question 1: In given circuit diagram two resistors are connected to a 5V supply. i. Calculate potential difference across the 8Q resistor. ii. A third resistor is now connected in parallel wit

Accepted Answer

Answer: Total current flowing through the circuit, I = = = = 0.36 A ∴ Potential difference across 8 f2 (Vi) = 0.36 × 8 = 2.88 V

Question 10

Question 2: Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.

Accepted Answer

i. Consider a part of conducting wire with its free electrons having the drift speed vd in the direction opposite to the electric field .

Question 11

Question 1: Distinguish between ohmic and non ohmic substances; explain with the help of example.

Accepted Answer

Step-by-step solution provided.

Question 12

Question 2: DC current flows in a metal piece of non uniform cross-section. Which of these quantities remains constant along the conductor: current, current density or drift speed?

Accepted Answer

Drift velocity and current density will change as it depends upon area of cross-section whereas current will remain constant.

Question 13

Question 1: What is the resistance of one of the rails of a railway track 20 km long at 20°C? The cross-section area of rail is 25 cm² and the rail is made of steel having resistivity at 20°C

Accepted Answer

Given: l = 20 km = 20 × 10³ m, A = 25 cm² = 25 × 10-4 m², ρ = 6 × 10-8 Ω m To find: Resistance of rail (R) Formula: ρ = Calculation: From formula. R = ρ ∴ R = = × 10-1 = 0.48 Ω

Question 14

Question 2: A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery. Can this battery drive starting motor of car?

Accepted Answer

E = 24 V, r = 380 Ω i. Maximum current (Imax) ii. Can battery start the motor? Formula: Imax = Calculation: From formula, Imax = = 0,063 A As, the value of current is very small compared to required current to run a starting motor of a car, this battery cannot be used to drive the motor.

Question 15

Question 3: A battery of emf 12 V and internal resistance 3 O is connected to a resistor. If the current in the circuit is 0.5 A, i. Calculate resistance of resistor. ii. Calculate terminal vo

Accepted Answer

Given: E = 12 V, r = 3 Ω, I = 0.5 A To find: i. Resistance (R) ii. Terminal voltage (V) Formulae: i. E = I (r + R) ii. V = IR Calculation: From formula (i), E = Ir + IR ∴ R = = = 21 Ω From formula (ii), V = 0.5 × 21 = 10.5 V

Question 16

Question 4: The magnitude of current density in a copper wire is 500 A/cm². If the number of free electrons per cm³ of copper is 8.47 × 1022, calculate the drift velocity of the electrons thro

Accepted Answer

Given: J = 500 A/cm² = 500 × 104 A/m², n = 8.47 × 1022 electrons/cm³ = 8.47 × 1028 electrons/m³ e = 1.6 × 10-19 C To Find: Drift velocity (vd) Formula: vd = Calculation: From formula, vd = = × 10-5 = {antilog [log 500 – log 8.47 – log 1.6]} × 10-5 = {antilog [2.6990 – 0.9279 -0.2041]} × 10-5 = {anti

Question 17

Question 5: Three resistors 10 Ω, 20 Ω and 30 Ω are connected in series combination. i. Find equivalent resistance of series combination. ii. When this series combination is connected to 12 V

Accepted Answer

Given: R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω, V = 12 V To Find: i. Series equivalent resistance(Rs) ii. Potential difference across each resistor (V1, V2, V3) Formula: i. Rs = R1 + R2 + R3 ii. V = IR Calculation: From formula (i), Rs = 10 + 20 + 30 = 60 Ω From formula (ii), I = = = 0.2 A ∴ Potential diffe

Question 18

Question 6: Two resistors 1 Ω and 2 Ω are connected in parallel combination. i. Find equivalent resistance of parallel combination. ii. When this parallel combination is connected to 9 V suppl

Accepted Answer

R1 = 1 kΩ = 10³ Ω, R2 = 2 kΩ = 2 × 10³ Ω, V = 9 V To find: i. Parallel equivalent resistance (Rp) ii. Current through 1 kΩ and 2 kΩ (I1 and I2) Formula: i. = + ii. V = IR Calculation: From formula (i), = + = ∴ Rp = = 0.66 kΩ From formula (ii), I1 = + = 9 × 10-3 A = 3 mA I2 = + = 4.5 × 10-3 A = 4.5 m

Question 19

Question 7: A silver wire has a resistance of 4.2 Ω at 27°C and resistance 5.4 Ω at 100°C. Determine the temperature coefficient of resistance.

Accepted Answer

Given: R1 =4.2 Ω, R2 = 5.4 Ω, T, = 27° C, T2= 100 °C To find: Temperature coefficient of resistance (α) Formula: α = Calculation: From Formula α = = 3.91 × 10-3/°C

Question 20

Question 8: A 6 m long wire has diameter 0.5 mm. Its resistance is 50 Ω. Find the resistivity and conductivity.

Accepted Answer

Given: l = 6 m, D = 0.5 mm, r = 0.25 mm = 0.25 × 10-3 m, R = 50 Ω To find: i. Resistivity (ρ) ii. Conductivity (σ) Formulae: i. ρ = = ii. σ = Calculation: From formula (i), ρ = = {antilog [log 50 + log 3.142 + 21og 0.25 -log 6]} × 10-6 = {antilog [ 1.6990 + 0.4972 + 2(1.3979) -0.7782]} × 10-6 = {ant

Question 21

Question 9: Find the value of resistances for the following colour code. i. Blue Green Red Gold ii. Brown Black Red Silver iii. Red Red Orange Gold iv. Orange White Red Gold v. Yellow Violet B

Accepted Answer

i. Given: Blue – Green – Red – Gold To find: Value of resistance Formula: Value of resistance = (xy × 10z ± T%)Ω Calculation:

Question 22

Question 10: Find the colour code for the following value of resistor having tolerance ± 10%. i. 330 Ω ii. 100 Ω iii. 47 kΩ iv. 160 Ω v. 1 kΩ

Accepted Answer

Step-by-step solution provided.

Question 23

Question 11: A current of 4 A flows through an automobile headlight. How many electrons flow through the headlight in a time of 2 hrs?

Accepted Answer

Given: I = 4 A, t = 2 hrs = 2 × 60 × 60 s To find: Number of electrons (N) Formula: I = = Calculation: As we know, e = 1.6 × 10-19 C From formula, N = = = 18 × 10-23

Question 24

Question 12: The heating element connected to 230 V draws a current of 5 A. Determine the amount of heat dissipated in 1 hour (J = 4.2 J/cal).

Accepted Answer

Given: V = 230 V, I = 5 A, At = 1 hour = 60 × 60 sec To find: Heat dissipated (H) Formula: H = ∆U = I∆tV Calculation: From formula, H = 5 × 60 × 60 × 230 = 4.14 × 106 J Heat dissipated in calorie, H = = 985.7 × 10³ cal = 985.7 kcal

Ohmic substances	Non-ohmic substances
1. Substances which obey ohm’s law are called ohmic substances.	Substances which do not obey ohm’s law are called non-ohmic substances.
2. Potential difference (V) versus current (I) curve is a straight line.	Potential difference (V) versus current (I) curve is not a straight line.
3. Resistance of these substances is constant i.e. they follow linear I-V characteristic.	Resistance of these substances
Expression for resistance is, R =	Expression for resistance is, R =
Examples: Gold, silver, copper etc.	Examples: Liquid electrolytes, vacuum tubes, junction diodes, thermistors etc.

Colour	Blue (x)	Green (y)	Red (z)	Gold (T%)
Code	6	5	2	± 5

Colour	Brown (x)	Black (y)	Red (z)	sliver (T%)
Code	1	0	2	± 10

Colour	Red (x)	Red (y)	Orange (z)	Gold (T%)
Code	2	2	3	± 5

Colour	Ornage (x)	White (y)	Red (z)	Gold (T%)
Code	3	9	2	± 5

Colour	Yellow (x)	violet (y)	Brown (z)	Sliver (T%)
Code	4	7	1	± 10

Chapter 11 Electric Current Through Conductors Solutions