Electric Current Archives

81. A current of 1·0 A is drawn by a filament of an electric bulb for 10 m

A current of 1·0 A is drawn by a filament of an electric bulb for 10 minutes. The amount of electric charge that flows through the circuit is

[amp_mcq option1=”0·1 C” option2=”10 C” option3=”600 C” option4=”800 C” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2021

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The correct answer is C) 600 C. The amount of electric charge (Q) that flows through a circuit is given by the product of the current (I) and the time (t), i.e., Q = I × t.

– Current (I) is given as 1.0 A.
– Time (t) is given as 10 minutes.
– Time must be converted to seconds for the calculation: 10 minutes * 60 seconds/minute = 600 seconds.
– Q = 1.0 A * 600 s = 600 Coulombs.

– The unit of electric charge is the Coulomb (C).
– The unit of electric current is the Ampere (A), which is defined as one Coulomb per second (C/s).

82. LED (a semi-conductor device) is an abbreviation that stands for

LED (a semi-conductor device) is an abbreviation that stands for

[amp_mcq option1=”Licence for Energy Detector.” option2=”Light Energy Device.” option3=”Light Emitting Diode.” option4=”Lost Energy Detector.” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2021

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LED stands for Light Emitting Diode.

A Light Emitting Diode (LED) is a semiconductor device that emits light when an electric current passes through it. This effect is a form of electroluminescence.

LEDs are widely used in various applications, including indicator lights, displays, lighting, and backlighting for screens, due to their energy efficiency, longevity, and compact size.

83. A fuse wire must be

A fuse wire must be

[amp_mcq option1=”conducting and of low melting point” option2=”conducting and of high melting point” option3=”insulator and of high melting point” option4=”insulator and of low melting point” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2019

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The correct option is A) conducting and of low melting point. A fuse wire must possess these properties to function correctly as a safety device.

– A fuse wire is designed to protect electrical circuits and appliances from damage due to excessive current.
– It is connected in series with the circuit, so it must be able to conduct electricity under normal operating conditions. Thus, it must be **conducting**.
– When the current flowing through the circuit exceeds a specified limit (the fuse rating), the fuse wire heats up due to the Joule effect (heat = I²Rt).
– To break the circuit and stop the excessive current flow, the fuse wire must melt when it gets too hot. This requires it to have a **low melting point**.
– If the fuse wire had a high melting point or was an insulator, it would not serve its purpose as a safety device.

Fuse wires are typically made of materials like tin, lead, or an alloy of tin and lead, which have relatively low melting points compared to the wires used in the rest of the circuit (like copper). The thickness and material of the fuse wire are carefully chosen to ensure it melts at the desired current level. Modern circuits often use circuit breakers instead of fuses, which are resettable switches that automatically trip when current is too high.

84. Consider the following statements about a solenoid: 1. The magnetic

Consider the following statements about a solenoid:

1. The magnetic field strength in a solenoid depends upon the number of turns per unit length in the solenoid
2. The magnetic field strength in a solenoid depends upon the current flowing in the wire of the solenoid
3. The magnetic field strength in a solenoid depends upon the diameter of the solenoid

Which of the statements given above are correct ?

[amp_mcq option1=”1, 2 and 3″ option2=”1 and 3 only” option3=”2 and 3 only” option4=”1 and 2 only” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2019

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Statements 1 and 2 are correct.

The magnetic field strength inside a long solenoid is given by the formula B ≈ μ₀ * n * I, where μ₀ is the permeability of free space, n is the number of turns per unit length (n = N/L), and I is the current flowing through the solenoid.
Statement 1 is correct as B is directly proportional to n (number of turns per unit length).
Statement 2 is correct as B is directly proportional to I (current).
Statement 3 is incorrect. For an ideal long solenoid, the magnetic field strength inside the solenoid (away from the ends) does not depend on its diameter. While the diameter might have minor effects in non-ideal cases (finite length solenoid, field near ends), the standard formula and theoretical treatment indicate no dependence on diameter for the uniform field inside a long solenoid.

The formula B = μ₀nI is an approximation that holds well for solenoids whose length is much greater than their diameter. In such cases, the magnetic field is strong, uniform inside (except near the ends), and zero outside.

85. Two metallic wires A and B are made using copper. The radius of wire A

Two metallic wires A and B are made using copper. The radius of wire A is r while its length is l. A dc voltage V is applied across the wire A, causing power dissipation, P. The radius of wire B is 2r and its length is 2l and the same dc voltage V is applied across it causing power dissipation, P₁. Which one of the following is the correct relationship between P and P₁?

[amp_mcq option1=”P = 2P₁” option2=”P = P₁/2″ option3=”P = 4P₁” option4=”P = P₁” correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2019

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The correct relationship is P = P₁/2.

The resistance of a wire is given by R = ρ * (l / A), where ρ is resistivity, l is length, and A is the cross-sectional area (A = πr²). Power dissipated is P = V²/R.
For wire A: R_A = ρ * (l / (πr²)). Power P = V² / R_A = V² / [ρ * (l / (πr²))] = (V²πr²) / (ρl).
For wire B: radius is 2r, length is 2l. R_B = ρ * (2l / (π(2r)²)) = ρ * (2l / (4πr²)) = ρ * (l / (2πr²)).
Power P₁ = V² / R_B = V² / [ρ * (l / (2πr²))] = (2V²πr²) / (ρl).
Comparing P and P₁: P₁ = 2 * [(V²πr²) / (ρl)].
Since P = (V²πr²) / (ρl), we have P₁ = 2P. This relationship can be rewritten as P = P₁/2.

The resistance of wire B is half the resistance of wire A (R_B = (ρl / (2πr²)) compared to R_A = (ρl / (πr²))). Since power is inversely proportional to resistance for a constant voltage (P = V²/R), lower resistance leads to higher power dissipation. Therefore, wire B dissipates more power than wire A when the same voltage is applied.

86. Let us consider a copper wire having radius r and length l. Let its re

Let us consider a copper wire having radius r and length l. Let its resistance be R. If the radius of another copper wire is 2r and the length is l/2 then the resistance of this wire will be

[amp_mcq option1=”R” option2=”2R” option3=”R/4″ option4=”R/8″ correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2019

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The correct answer is (D) R/8.

The resistance (R) of a wire is directly proportional to its length (l) and inversely proportional to its cross-sectional area (A). The formula is R = ρ(l/A), where ρ is the resistivity of the material. The cross-sectional area of a wire with radius r is A = πr².

Let the initial resistance be R. R = ρ * (l / (πr²)).
For the second wire, the radius is r’ = 2r and the length is l’ = l/2.
The new cross-sectional area A’ = π(r’)² = π(2r)² = π(4r²) = 4πr².
The new resistance R’ = ρ * (l’ / A’) = ρ * ((l/2) / (4πr²)).
R’ = ρ * (l / (8πr²)) = (1/8) * [ρ * (l / (πr²))].
Since R = ρ * (l / (πr²)), we have R’ = R/8.

87. Which one of the following statements about magnetic field lines is NO

Which one of the following statements about magnetic field lines is NOT correct?

[amp_mcq option1=”They can emanate from a point” option2=”They do not cross each other” option3=”Field lines between two poles cannot be precisely straight lines at the ends” option4=”There are no field lines within a bar magnet” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2018

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Magnetic field lines form continuous closed loops. Outside a magnet, they are depicted as originating from the North pole and terminating at the South pole. Inside the magnet, the field lines continue from the South pole to the North pole, completing the loop. Therefore, there *are* magnetic field lines within a bar magnet.

Magnetic field lines are continuous and form closed loops, unlike electric field lines which start on positive charges and end on negative charges. This reflects the absence of magnetic monopoles.

Statement A is broadly correct in that lines emanate from the pole regions (which are not mathematical points but areas) of a magnet. Statement B is a fundamental property: field lines never cross each other, as this would imply two different directions for the magnetic field at a single point, which is impossible. Statement C is correct: field lines are generally curved, especially near the poles and in the region between them, only becoming approximately straight far from the magnet or in very specific configurations.

88. Which one of the following devices is non-ohmic ?

Which one of the following devices is non-ohmic ?

[amp_mcq option1=”Conducting copper coil” option2=”Electric heating coil” option3=”Semi conductor diode” option4=”Rheostat” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2018

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An ohmic device obeys Ohm’s Law (V = IR) where resistance (R) is constant regardless of voltage (V) or current (I). A non-ohmic device does not obey Ohm’s Law; its resistance varies with the applied voltage or current.
– Conducting copper coil (A) and Electric heating coil (B) (like Nichrome wire) are generally considered ohmic conductors, although their resistance can change with temperature.
– A Rheostat (D) is a variable resistor, but for any fixed resistance setting, it acts as an ohmic resistor.
– A Semiconductor diode (C) is a device that exhibits a highly non-linear current-voltage characteristic. It conducts current effectively in only one direction (forward bias) and only above a certain threshold voltage. Its resistance changes dramatically depending on the applied voltage and direction. Therefore, it is a classic example of a non-ohmic device.

– Ohmic devices have constant resistance independent of voltage or current.
– Non-ohmic devices have resistance that varies with voltage or current.
– Metal conductors are typically ohmic (assuming constant temperature).
– Semiconductor devices like diodes and transistors are typically non-ohmic.

The V-I graph for an ohmic device is a straight line passing through the origin with a slope equal to the resistance. The V-I graph for a non-ohmic device is curved. Examples of non-ohmic devices include vacuum tubes, gas discharge lamps, and semiconductor components like diodes and transistors.

89. Suppose a rod is given a negative charge by rubbing it with wool. Whic

Suppose a rod is given a negative charge by rubbing it with wool. Which one of the following statements is correct in this case ?

[amp_mcq option1=”The positive charges are transferred from rod to wool” option2=”The positive charges are transferred from wool to rod” option3=”The negative charges are transferred from rod to wool” option4=”The negative charges are transferred from wool to rod” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2017

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The question describes a rod being given a negative charge by rubbing it with wool and asks about the charge transfer process.

When two different materials are rubbed together, electrons can be transferred from one material to the other due to differences in their electron affinities. This process is called charging by friction or triboelectric effect. If the rod gains a negative charge, it means it has received negative charge carriers, which are electrons. Since positive charges (protons) are fixed within the nucleus of atoms in the solid material, they are generally not transferred during rubbing. Therefore, the negative charge acquired by the rod must come from the wool.
– If the rod becomes negatively charged, it has *gained* electrons.
– These electrons must have come from the wool.
So, negative charges (electrons) are transferred from the wool to the rod. The wool loses electrons and becomes positively charged.

The triboelectric series lists materials in order of their tendency to gain or lose electrons when rubbed. Materials higher on the list tend to lose electrons and become positively charged, while materials lower on the list tend to gain electrons and become negatively charged. Wool is typically higher than many common materials like plastic or ebonite (used for rods), indicating it tends to lose electrons to these materials when rubbed.

90. Which one of the following physical quantities does NOT affect the res

Which one of the following physical quantities does NOT affect the resistance of a cylindrical resistor ?

[amp_mcq option1=”The current through it” option2=”Its length” option3=”The resistivity of the material used in the resistor” option4=”The area of cross-section of the cylinder” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2017

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The current through the cylindrical resistor does NOT affect its resistance.

The resistance of a resistor is a property of the material and its geometry. For a cylindrical resistor made of a homogeneous material, its resistance (R) is determined by the formula R = ρ * (L/A), where ρ (rho) is the resistivity of the material, L is the length of the resistor, and A is the area of its cross-section.

According to the formula R = ρ * (L/A), the resistance is affected by the resistivity of the material, the length, and the area of cross-section. The current flowing through the resistor is a result of the applied voltage and the resistor’s resistance (as described by Ohm’s law, V=IR). Changing the voltage changes the current, but the resistance of the resistor itself remains constant (assuming constant temperature and material properties). Therefore, the current through the resistor does not affect its resistance; rather, the resistance affects the current for a given voltage.