308 + Interview Questions in ELECTRIC CURRENT IN GENERAL AWARENESS

1.	Which of the cables, one rated 5A and other 15 A will be thicker wire? Give reason for your answer.
Answer» i) To carry larger current, the resistance of wire should be low, so its area of cross-section should be large. ii) Because the second cable carrying 15 A ampere current which indicates it has low resistance that is more surface area which implies it is thicker wire.

1.

Which of the cables, one rated 5A and other 15 A will be thicker wire? Give reason for your answer.

Answer»

i) To carry larger current, the resistance of wire should be low, so its area of cross-section should be large.

ii) Because the second cable carrying 15 A ampere current which indicates it has low resistance that is more surface area which implies it is thicker wire.

Discussion

2.	The V-I graph for a metal conductor will be …………………. when provided the temperature and other physical conditions remain constant A) non-linear B) linear C) parabola D) curved graph
Answer» Correct option is B) linear

Discussion

3.	Three resistors A, B and C are connected as shown in the figure. Each of them dissipates energy to a maximum of 18 W. Find the maximum current that can flow through the three resistors.
Answer» Resultant Resistance in the circuit = 2Ω + 1/ 2Ω + 1/2Ω = 2 + (1) = 3Ω P = i² R i = √P/R = √18W/3Ω = √6 AMP

3.

Three resistors A, B and C are connected as shown in the figure. Each of them dissipates energy to a maximum of 18 W. Find the maximum current that can flow through the three resistors.

Answer»

Resultant Resistance in the circuit = 2Ω + 1/ 2Ω + 1/2Ω

= 2 + (1) = 3Ω

P = i² R

i = √P/R = √18W/3Ω = √6 AMP

Discussion

4.	Explain the conditions for maximum current.
Answer» 1. Current in a circuit is given by, I = \(\frac{E}{R+r}\) 2. Maximum current can be drawn from the emf device, only when R = 0, i.e. I_max = \(\frac{E}{R}\) 3. I_max is the maximum allowed current from an emf device (or a cell) which decides the maximum current rating of a cell or a battery.

4.

Explain the conditions for maximum current.

Answer»

1. Current in a circuit is given by, I = \(\frac{E}{R+r}\)

2. Maximum current can be drawn from the emf device, only when R = 0, i.e.

I_max = \(\frac{E}{R}\)

3. I_max is the maximum allowed current from an emf device (or a cell) which decides the maximum current rating of a cell or a battery.

Discussion

5.	What happens if we use a fuse made up of same wire which is used to make the electric circuit?
Answer» It doesn’t work as a fuse. If high voltage occurs fuse do not melt and circuit will not be opened / breaked. So home appliances will be damaged.

Discussion

6.	What is the reason for connecting the fuse in the live wire?
Answer» 1. The fuse is always connected in the live wire of the circuit. 2. If the fuse is put in the neutral wire, due to excessive flow of current the fuse burns, current stops flowing in the circuit. 3. But the appliance remains connected to the high potential point of the supply through the live wire. 4. Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

6.

What is the reason for connecting the fuse in the live wire?

Answer»

1. The fuse is always connected in the live wire of the circuit.

2. If the fuse is put in the neutral wire, due to excessive flow of current the fuse burns, current stops flowing in the circuit.

3. But the appliance remains connected to the high potential point of the supply through the live wire.

4. Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

Discussion

7.	Why is the fuse wire fitted in a porcelain casing?
Answer» Porcelain is an insulator of electricity. So fuse wire is fitted in a porcelain casing.

Discussion

8.	Three resistors each of value 3 Ω are connected, in parallel combination. Their equivalent resistance is ……………………. A) 9 Ω B) 1 Ω C) 0.33 Ω D) 1.5 Ω
Answer» Correct option is B) 1 Ω

Discussion

9.	Three resistors of values 2 Ω, 4 Ω, 6 Ω are connected in parallel. The equivalent resistance of combination of resistors is ………………… A) 6 Ω B) 1 ΩC) 1.09 Ω D) 12 Ω
Answer» Correct option is C) 1.09 Ω

Discussion

10.	Observe the above diagram and answer the following. a) Are all the resistors connected in parallel or series? b) What is the equivalent resistance of the combination of three resistors? c) In this system, which physical quantity is constant? d) If R1 = 2Ω, R2 = 3Ω, R3 = 4Ω. find equivalent resistance.
Answer» a) Connected in series. b) R_eq = R₁ + R₂ + R₃ c) Current (I) d) R_eq = R₁ + R₂ + R₃ =2 + 3 + 4 = 9Ω

10.

Observe the above diagram and answer the following. a) Are all the resistors connected in parallel or series? b) What is the equivalent resistance of the combination of three resistors? c) In this system, which physical quantity is constant? d) If R1 = 2Ω, R2 = 3Ω, R3 = 4Ω. find equivalent resistance.

Answer»

a) Connected in series.

b) R_eq = R₁ + R₂ + R₃

c) Current (I)

d) R_eq = R₁ + R₂ + R₃

=2 + 3 + 4 = 9Ω

Discussion

11.	Why are copper wires used as connecting wires?
Answer» Copper is a good conductor of electricity so copper wires are used as connecting wires.

Discussion

12.	How can you appreciate the role of a small fuse in house wiring circuit in preventing damage to various electrical appliances connected in the circuit? (OR) We can save the household wiring and devices by using fuses. Write any four points by appreciating the role of fuse.
Answer» The fuse consists of a thin wire of low melting point. When the current in the fuse exceeds 20 A, the wire will heat up and melt. The circuit then becomes open and prevents the flow of current into the household circuit. So all the electric devices are saved from damage that could be caused by overload. Thus we can save the household wiring and devices by using fuses. In this way a small fuse prevents a great damage to costly electrical appliances in the circuit.

Discussion

13.	Why don’t we use series arrangement of electrical appliances like bulb, television, fan, and others in domestic circuits?
Answer» 1. If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases. To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series. They are connected in parallel. 2. In series combination same current passes through all resistors. This is not suggestable for household appliances. Hence, they are connected in parallel.

13.

Why don’t we use series arrangement of electrical appliances like bulb, television, fan, and others in domestic circuits?

Answer»

1. If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases. To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series. They are connected in parallel.

2. In series combination same current passes through all resistors. This is not suggestable for household appliances. Hence, they are connected in parallel.

Discussion

14.	Overloading of household circuit means ………………….. A) Increasing the number of electric appliances consuming power more than 4.8 KW B) The household circuits draw the current from mains more than 20 A C) The household circuits draw the current from mains more than 40 A D) Both A and B
Answer» D) Both A and B

Discussion

15.	Which of the following is used to prevent damages due to overloading to the household circuit ? A) switch B) fuseC) resistanceD) all
Answer» Correct option is B) fuse

Discussion

16.	State whether the home appliances like Fridge, TV, Computer are connected in series or parallel Why?
Answer» They are connected in parallel because if any one device is damaged, the rest will work as usual because the circuit does not break.

Discussion

17.	What happens when this current (overloading) increases greatly to the household circuit?
Answer» It causes fire.

Discussion

18.	How much current is drawn from the battery if the resistors are connected in parallel? Is it equal to individual currents drawn by the resistors?
Answer» It is the sum of currents flowing through each resistor. No, it is the sum of individual currents drawn by the resistors.

Discussion

19.	What happens when one of the resistors in series breaks down?
Answer» The circuit becomes open and flow of current will be broken down.

Discussion

20.	Why should we connect electric appliances in parallel in a household circuit? What happens if they are connected in series?
Answer» The electric appliances are connected in parallel in a household circuit. Because in parallel wiring if any electric appliance is switched off, other appliances don’t get off. If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases. To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series.

Discussion

21.	Statement X : When resistances are connected in parallel the reciprocal of equivalent resistance is equal to the sum of reciprocal of their individual resistances. Statement Y : In parallel combination of resistors the potential difference across all the resistors is same.A) Both statements are true B) Both statements are false C) X is true, Y is false D) X is false, Y is true.
Answer» A) Both statements are true

Discussion

22.	Three resistors of resistances 3 Ω, 4 Ω and 6 Ω are connected in parallel. The combination is connected to a battery of 8 V. The current in the circuit is …………………A) 4 A B) 3 A C) 6 A D) 8 A
Answer» Correct option is C) 6 A

Discussion

23.	Three equal resistances are connected In series, then in parallel. What will be the ratio of their resultant resistances?
Answer» Suppose the resistance of equal resistors is ‘R’. Suppose they are connected in series. Then their equivalent resistance R’ = R + R + R = 3R If they are connected in parallel their equivalent resistance R” = R/3 ∴ Ratio of resultant resistances = R’: R” = 3R : R/3 = 9:1

23.

Three equal resistances are connected In series, then in parallel. What will be the ratio of their resultant resistances?

Answer»

Suppose the resistance of equal resistors is ‘R’. Suppose they are connected in series.

Then their equivalent resistance R’ = R + R + R = 3R

If they are connected in parallel their equivalent resistance R” = R/3

∴ Ratio of resultant resistances = R’: R”

= 3R : R/3 = 9:1

Discussion

24.	How is Ideal earthing helpful during short circuiting?
Answer» 1. During short circuiting an excessive current flows through the live wires. 2. It will pass to earth through the earth wire if there is local earthing otherwise it may cause a fire due to overheating of the live wires.

24.

How is Ideal earthing helpful during short circuiting?

Answer»

1. During short circuiting an excessive current flows through the live wires.

2. It will pass to earth through the earth wire if there is local earthing otherwise it may cause a fire due to overheating of the live wires.

Discussion

25.	The resistance of a conductor …………………. when the length of a conductor increased for providing constant temperature and cross-sectional area. A) decreases B) increases C) remains same D) none
Answer» B) increases

Discussion

26.	The S.I unit of resistance is …………………… A) volt B) ohm C) ampere D) coloumb
Answer» Correct option is B) ohm

Discussion

27.	Statement X : The resistance of a conductor is inversely proportional to its cross sectional area. Statement Y : The resistance of a wire is inversely proportional to the square of its diameter. A) Both statements are true B) Both statements are false C) X is true, Y is false D) X is false, Y is true
Answer» A) Both statements are true

Discussion

28.	Which of the following is an equation for Ohm’s law ? A) V = IRB) I = V/RC) R = V/ID) All
Answer» Correct option is D) All

Discussion

29.	The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature. This law is known as ……………. A) Ohm’s lawB) Joule’s law C) Power’s law D) Faraday’s law
Answer» A) Ohm’s law

Discussion

30.	If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance will be(A) 0.1% (B) 0.2% (C) 1 % (D) 2 %
Answer» Correct option is: (C) 1 %

Discussion

31.

Evaluate resistance for the following colour coded resistors:i. Yellow – Violet – Black – Silver ii. Green – Blue – Red – Gold iii. Brown – Black – Orange – Gold

Answer»

i. Given: Yellow – Violet – Black – Silver

To find: Value of resistance

Formula: Value of resistance

= (xy × 10^z ± T%)Ω

Colour	Yellow (x)	Violet (y)	Black (z)	Sliver (T%)
Code	4	7	0	±10

Hence x = 4, y = 7, z = 0, T = 10%

Value of resistance = (xy ×10^z ± T%) Ω

= (47 × 10° ± 10%) Ω

Value of resistance = 47 Ω ± 10%

To find: Value of resistance

Formula: Value of resistance

= (xy × 10^z ± T%) Ω

Calculation:

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

Hence x = 5, y = 6, z = 2, T = 5%

Value of resistance = (xy × 10^z ± T%) Q

= 56 × 102 Ω ± 5%

= 5.6 k Ω ± 5%

iii. Given: Brown – Black – Orange – Gold

To find: Value of the resistance

Formula: Value of the resistance

= (xy × 10^z ± T%) Ω

Calculation:

Colour	Brown (x)	Black (y)	Orange (z)	Gold (T%)
Code	1	0	3	±5

Hence x = 1, y = 0, z = 3, T = 5%

Value of resistance = (xy × 10^z ± T%) Ω

= 10 × 10³ Ω ± 5%

= 10 kΩ ± 5%

Discussion

32.

Colour code of resistor is Yellow-Violet Orange-Gold. Find its value.

Answer»

	Yellow (x)	Violet (y)	Orange (z)	Gold (T%)
Value	4	7	3	± 5

Value of resistance: xy × 10^z Ω ± tolerance

∴ Value of resistance = 47 × 10³ Ω ± 5%

= 47 kΩ ± 5%

Discussion

33.	Three resistors, 8 Ω, 4 Ω and 10 Ω connected in parallel as shown in figure, the equivalent resistance is(A) \(\frac{19}{40}\) Ω(B) \(\frac{40}{19}\) Ω(C) \(\frac{80}{19}\) Ω (D) \(\frac{34}{23}\) Ω
Answer» Correct option is: (B) \(\frac{40}{19}\) Ω

Discussion

34.	Find the value of current which is flowing through the bulb of 40 W 220 V.
Answer» P = VI ⇒ I = \(\frac{P}{V}=\frac{40}{220}\) = 0.1818 amp

34.

Find the value of current which is flowing through the bulb of 40 W 220 V.

Answer»

P = VI

⇒ I = \(\frac{P}{V}=\frac{40}{220}\)

= 0.1818 amp

Discussion

35.	Write the S.I. unit of current density?
Answer» Current density, J = \(\frac{I}{A}\) = Ampere/metre²

Discussion

36.	What will be the value of charge when an electric current flows through a conductor?
Answer» On flowing a electric current in conductor it does not charged. So total charge across it will be zero.

Discussion

37.	A conducting resistance is connected to the battery and temperature of conductor decreases by the process of cooling then the value of current will be:(a) increased(b) decreased(c) remain constant(d) zero
Answer» (a) increased The resistance of the conducting wire will decrease according to equation R_t = R₀ (1 ± α Δt) due to which current will increase according to Ohm’s law (I = \(\frac{V}{R}\))

Discussion

38.	When the wire is connected with a battery is heated up due to electric current which quantity do not vary:(a) Drift velocity(b) Resistivity(c) Resistance(d) Number of free electrons
Answer» (d) Number of free electrons Number of free electrons does not depend on temperature.

Discussion

39.	A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6 × 10-7 m2, and its resistance is measured to be 5 Ω. What is the resistivity of the material at the temperature of the experiment?
Answer» Given: l = 15 m, A = 6.0 × 10^-7 m², R = 5 Ω To find: Resistivity (ρ) Formula: ρ = \(\frac{RA}{l}\) Calculation: From formula, ρ = \(\frac{5\times6\times10^{-7}}{15}\) ∴ ρ = 2 × 10^-7 Ω m

39.

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6 × 10-7 m2, and its resistance is measured to be 5 Ω. What is the resistivity of the material at the temperature of the experiment?

Answer»

Given: l = 15 m, A = 6.0 × 10^-7 m², R = 5 Ω

To find: Resistivity (ρ)

Formula: ρ = \(\frac{RA}{l}\)

Calculation: From formula,

ρ = \(\frac{5\times6\times10^{-7}}{15}\)

∴ ρ = 2 × 10^-7 Ω m

Discussion

40.	The terminal voltage across a cell is more than its e.m.f., if another cell of(A) higher e.m.f. is connected parallel to it. (B) less e.m.f. is connected parallel to it. (C) less e.m.f. is connected in series with it. (D) higher e.m.f. is connected in series with it.
Answer» (A) higher e.m.f. is connected parallel to it.

Discussion

41.	If n identical cells, each of emf E and internal resistance r, are connected in series, write an expression for the terminal p.d. of the combination and hence show that this is nearly n times that of a single cell.
Answer» i. Let n identical cells, each of emf E and internal resistance r, be connected in series. Let the current supplied by this combination to an external resistance R be I. ii. The equivalent emf of the combination, E_eq = E + E + …….. (n times) = nE iii. The equivalent internal resistance of the combination, r_eq = r + r + … (n times) = nr iv. The terminal p.d. of the combination is V = E_eq– Ir_eq = nE – Inr = n (E – Ir) ∴ V = n × terminal p.d. of a single cell Thus, the terminal p.d. of the series combination is n times that of a single cell.

41.

If n identical cells, each of emf E and internal resistance r, are connected in series, write an expression for the terminal p.d. of the combination and hence show that this is nearly n times that of a single cell.

Answer»

i. Let n identical cells, each of emf E and internal resistance r, be connected in series. Let the current supplied by this combination to an external resistance R be I.

ii. The equivalent emf of the combination, E_eq = E + E + …….. (n times) = nE

iii. The equivalent internal resistance of the combination,

r_eq = r + r + … (n times)

= nr

iv. The terminal p.d. of the combination is

V = E_eq– Ir_eq = nE – Inr = n (E – Ir)

∴ V = n × terminal p.d. of a single cell

Thus, the terminal p.d. of the series combination is n times that of a single cell.

Discussion

42.	Statement X : emf of a battery is taken as positive when we move from negative terminal to positive terminal across the battery in a closed circuit loop. Statement Y : emf of a battery is taken as positive when we move from positive terminal to negative terminal across the battery in a closed circuit loop. A) Both statements are true B) Both statements are false C) X is true, Y is false D) X is false, Y is true
Answer» C) X is true, Y is false

Discussion

43.	The V-I graph for a series combination and for parallel combination of two resistors is shown in figure. Which of the two A or B, represent parallel combination? Give reason for your answer.
Answer» i) For same change in I, change in V is less for the straight line A than for the straight line B (i.e., straight line A is less steeper than B). ii) So the straight line A represents small resistance, while straight line B represents more resistance. iii) In parallel combination, the resistance decreases, while in series combination, the resistance increases, so A represents a parallel combination.

43.

The V-I graph for a series combination and for parallel combination of two resistors is shown in figure. Which of the two A or B, represent parallel combination? Give reason for your answer.

Answer»

i) For same change in I, change in V is less for the straight line A than for the straight line B (i.e., straight line A is less steeper than B).

ii) So the straight line A represents small resistance, while straight line B represents more resistance.

iii) In parallel combination, the resistance decreases, while in series combination, the resistance increases, so A represents a parallel combination.

Discussion

44.	Calculate:i. total resistance andii. total current in the following circuit.R1 = 3 Ω, R2 = 6 Ω, R3 = 5 Ω, V = 14 V
Answer» i. R₁ and R₂ are connected in parallel. This combination (R_P) is connected in series with R₃. ∴ Total resistance, R_T = R_P + R₃ R_P = \(\frac{R_1R_2}{R_1+R_2}=\frac{3\times6}{3+6}\) = 2 Ω ∴ R_T = 2 + 5 = 7 Ω ii. Total current: I = \(\frac{V}{R_T}=\frac{14}{7}\) = 2 A

44.

Calculate:i. total resistance andii. total current in the following circuit.R1 = 3 Ω, R2 = 6 Ω, R3 = 5 Ω, V = 14 V

Answer»

i. R₁ and R₂ are connected in parallel. This combination (R_P) is connected in series with R₃.

∴ Total resistance, R_T = R_P + R₃

R_P = \(\frac{R_1R_2}{R_1+R_2}=\frac{3\times6}{3+6}\) = 2 Ω

∴ R_T = 2 + 5 = 7 Ω

ii. Total current: I = \(\frac{V}{R_T}=\frac{14}{7}\) = 2 A

Discussion

45.	Name two special characteristics of heating coil.
Answer» High resistivity and high melting point.

Discussion

46.	How does resistivity vary with material of conductor?
Answer» The resistivity is less for a good conductor and is large for a bad conductor.

Discussion

47.	How does a battery maintain a constant potential difference between its terminals?
Answer» We know that a battery consists electric force (Fe) and chemical force (Fc). These two forces are balanced in a battery. Due to this reason a battery maintains a constant , potential difference between its terminals.

Discussion

48.	If a charge q is moving against an electric force Fe through a distance “d” between the terminals of a battery then the em/is …………………….. A) Fe .dq B) Fe/dq C)Fed/q D)Feq/d
Answer» Correct option is C)F_ed/q

Discussion

49.	An ammeter is always connected in ……………… to the circuit A) seriesB) parallel C) both series (or) parallel D) none
Answer» Correct option is A) series

Discussion

50.

If the resistance of your body is 100000 Cl, what would be the current that flows in your body when you touch the terminals of a 12 V battery?

Answer»

We know that, I = V/R; here V = 12 V, R = 1,00,000Ω.

∴ The current passing through our body

I = 12 V/100000Ω = 0.00012 Ampere.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Which of the cables, one rated 5A and other 15 A will be thicker wire? Give reason for your answer.

The V-I graph for a metal conductor will be …………………. when provided the temperature and other physical conditions remain constant A) non-linear B) linear C) parabola D) curved graph

Three resistors A, B and C are connected as shown in the figure. Each of them dissipates energy to a maximum of 18 W. Find the maximum current that can flow through the three resistors.

Explain the conditions for maximum current.

What happens if we use a fuse made up of same wire which is used to make the electric circuit?

What is the reason for connecting the fuse in the live wire?

Why is the fuse wire fitted in a porcelain casing?

Three resistors each of value 3 Ω are connected, in parallel combination. Their equivalent resistance is ……………………. A) 9 Ω B) 1 Ω C) 0.33 Ω D) 1.5 Ω

Three resistors of values 2 Ω, 4 Ω, 6 Ω are connected in parallel. The equivalent resistance of combination of resistors is ………………… A) 6 Ω B) 1 ΩC) 1.09 Ω D) 12 Ω

Why are copper wires used as connecting wires?

How can you appreciate the role of a small fuse in house wiring circuit in preventing damage to various electrical appliances connected in the circuit? (OR) We can save the household wiring and devices by using fuses. Write any four points by appreciating the role of fuse.

Why don’t we use series arrangement of electrical appliances like bulb, television, fan, and others in domestic circuits?

Overloading of household circuit means ………………….. A) Increasing the number of electric appliances consuming power more than 4.8 KW B) The household circuits draw the current from mains more than 20 A C) The household circuits draw the current from mains more than 40 A D) Both A and B

Which of the following is used to prevent damages due to overloading to the household circuit ? A) switch B) fuseC) resistanceD) all

State whether the home appliances like Fridge, TV, Computer are connected in series or parallel Why?

What happens when this current (overloading) increases greatly to the household circuit?

How much current is drawn from the battery if the resistors are connected in parallel? Is it equal to individual currents drawn by the resistors?

What happens when one of the resistors in series breaks down?

Why should we connect electric appliances in parallel in a household circuit? What happens if they are connected in series?

Three resistors of resistances 3 Ω, 4 Ω and 6 Ω are connected in parallel. The combination is connected to a battery of 8 V. The current in the circuit is …………………A) 4 A B) 3 A C) 6 A D) 8 A

Three equal resistances are connected In series, then in parallel. What will be the ratio of their resultant resistances?

How is Ideal earthing helpful during short circuiting?

The resistance of a conductor …………………. when the length of a conductor increased for providing constant temperature and cross-sectional area. A) decreases B) increases C) remains same D) none

The S.I unit of resistance is …………………… A) volt B) ohm C) ampere D) coloumb

Which of the following is an equation for Ohm’s law ? A) V = IRB) I = V/RC) R = V/ID) All

The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature. This law is known as ……………. A) Ohm’s lawB) Joule’s law C) Power’s law D) Faraday’s law

If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance will be(A) 0.1% (B) 0.2% (C) 1 % (D) 2 %

Evaluate resistance for the following colour coded resistors:i. Yellow – Violet – Black – Silver ii. Green – Blue – Red – Gold iii. Brown – Black – Orange – Gold

Colour code of resistor is Yellow-Violet Orange-Gold. Find its value.

Three resistors, 8 Ω, 4 Ω and 10 Ω connected in parallel as shown in figure, the equivalent resistance is(A) \(\frac{19}{40}\) Ω(B) \(\frac{40}{19}\) Ω(C) \(\frac{80}{19}\) Ω (D) \(\frac{34}{23}\) Ω

Find the value of current which is flowing through the bulb of 40 W 220 V.

Write the S.I. unit of current density?

What will be the value of charge when an electric current flows through a conductor?

A conducting resistance is connected to the battery and temperature of conductor decreases by the process of cooling then the value of current will be:(a) increased(b) decreased(c) remain constant(d) zero

When the wire is connected with a battery is heated up due to electric current which quantity do not vary:(a) Drift velocity(b) Resistivity(c) Resistance(d) Number of free electrons

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6 × 10-7 m2, and its resistance is measured to be 5 Ω. What is the resistivity of the material at the temperature of the experiment?

The terminal voltage across a cell is more than its e.m.f., if another cell of(A) higher e.m.f. is connected parallel to it. (B) less e.m.f. is connected parallel to it. (C) less e.m.f. is connected in series with it. (D) higher e.m.f. is connected in series with it.

If n identical cells, each of emf E and internal resistance r, are connected in series, write an expression for the terminal p.d. of the combination and hence show that this is nearly n times that of a single cell.

The V-I graph for a series combination and for parallel combination of two resistors is shown in figure. Which of the two A or B, represent parallel combination? Give reason for your answer.

Calculate:i. total resistance andii. total current in the following circuit.R1 = 3 Ω, R2 = 6 Ω, R3 = 5 Ω, V = 14 V

Name two special characteristics of heating coil.

How does resistivity vary with material of conductor?

How does a battery maintain a constant potential difference between its terminals?

If a charge q is moving against an electric force Fe through a distance “d” between the terminals of a battery then the em/is …………………….. A) Fe .dq B) Fe/dq C)Fed/q D)Feq/d

An ammeter is always connected in ……………… to the circuit A) seriesB) parallel C) both series (or) parallel D) none

If the resistance of your body is 100000 Cl, what would be the current that flows in your body when you touch the terminals of a 12 V battery?