Electric Current

51. Consider the following circuit : Which one of the following is the va

Consider the following circuit :

Which one of the following is the value of the resistance between points A and B in the circuit given above?

[amp_mcq option1=”2/5 R” option2=”3/5 R” option3=”3/2 R” option4=”4R” correct=”option2″]

This question was previously asked in
UPSC NDA-2 – 2018
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Based on external sources providing solutions for this specific problem diagram from UPSC exams, the equivalent resistance between points A and B is given as 3/5 R.
– The provided circuit diagram appears identical to a balanced Wheatstone bridge configuration where all resistors have the value R.
– Standard circuit analysis of a balanced Wheatstone bridge shows that the equivalent resistance between the input terminals (A and B in this configuration) is R.
– However, R is not among the given options. The value 3/5 R is a commonly cited answer for this specific problem diagram in certain exam contexts, suggesting a possible error in the question/options or an intended interpretation that deviates from standard balanced bridge analysis.
A standard analysis of this circuit as a balanced Wheatstone bridge (A-R-Node1, A-R-Node2, Node1-R-Node2, Node1-R-B, Node2-R-B) shows that the potential at Node1 equals the potential at Node2. Thus, no current flows through the central resistor (Node1-Node2). The circuit then simplifies to two parallel branches, each with resistance 2R (R+R), resulting in an equivalent resistance of (2R * 2R) / (2R + 2R) = R. Since this result is not in the options, and 3/5 R is frequently provided as the answer in the context of this question, there is a strong indication of an inconsistency or error in the problem statement or options as presented. However, adhering to the requirement of selecting from the provided options, 3/5 R (Option B) is selected based on external validation sources associated with this question.

52. If a free electron moves through a potential difference of 1 kV, then

If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by

[amp_mcq option1=”1.6 × 10⁻¹⁹ J” option2=”1.6 × 10⁻¹⁶ J” option3=”1 × 10⁻¹⁹ J” option4=”1 × 10⁻¹⁶ J” correct=”option2″]

This question was previously asked in
UPSC NDA-2 – 2018
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The energy gained or lost by a charge (q) moving through a potential difference (V) is given by ΔE = qV. The charge of a free electron is e = 1.602 × 10⁻¹⁹ Coulombs. The potential difference is 1 kV = 1000 Volts. Therefore, the energy gained is ΔE = (1.602 × 10⁻¹⁹ C) × (1000 V) = 1.602 × 10⁻¹⁶ Joules. Option B is the closest value.
This question tests the concept of electric potential energy and its relation to potential difference and charge (ΔE = qΔV). It also requires knowing the charge of an electron and unit conversions (kV to V).
The energy gained by an electron moving through a potential difference of 1 Volt is defined as 1 electronvolt (eV). So, moving through 1 kV (1000 V) gives an energy gain of 1000 eV. Since 1 eV = 1.602 × 10⁻¹⁹ J, 1000 eV = 1000 × 1.602 × 10⁻¹⁹ J = 1.602 × 10⁻¹⁶ J.

53. The magnetic field strength of a current-carrying wire at a particular

The magnetic field strength of a current-carrying wire at a particular distance from the axis of the wire

[amp_mcq option1=”depends upon the current in the wire” option2=”depends upon the radius of the wire” option3=”depends upon the temperature of the surroundings” option4=”None of the above” correct=”option1″]

This question was previously asked in
UPSC NDA-2 – 2018
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The magnetic field strength of a current-carrying wire at a particular distance from the axis of the wire depends upon the current in the wire.
– The magnetic field strength (B) around a long straight current-carrying wire is given by the formula B = (μ₀ * I) / (2π * r).
– In this formula, I represents the current in the wire, and r represents the perpendicular distance from the axis of the wire.
– The magnetic field strength is directly proportional to the current (I).
– The magnetic field strength is inversely proportional to the distance (r).
μ₀ is the permeability of free space, a constant. The formula shows the direct dependence on current and distance, but not explicitly on the wire’s radius or surrounding temperature in the basic model for an external point. Temperature could indirectly affect the magnetic field by influencing the wire’s resistance and thus the current (if voltage is constant), but the fundamental relationship is B ∝ I for a given geometry.

54. Step-up transformers are used for

Step-up transformers are used for

[amp_mcq option1=”increasing electrical power” option2=”decreasing electrical power” option3=”decreasing voltage” option4=”increasing voltage” correct=”option4″]

This question was previously asked in
UPSC NDA-2 – 2017
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A transformer is an electrical device that transfers electrical energy between two or more circuits through electromagnetic induction. Step-up transformers are specifically designed to increase (step up) the voltage from the primary coil to the secondary coil. They have more turns in the secondary coil than in the primary coil. Conversely, a step-down transformer decreases the voltage. Transformers are highly efficient, but they do not increase or decrease the total electrical power (ideally, power input equals power output, although small losses occur in real transformers).
– Step-up transformers increase voltage.
– They work based on electromagnetic induction.
– They are used in AC circuits.
Step-up transformers are commonly used in power transmission systems. Electricity is generated at a relatively low voltage at power plants. Step-up transformers increase the voltage to very high levels for transmission over long distances (reducing current and thus minimizing power loss due to resistance, P = I²R). At the receiving end, step-down transformers are used to reduce the voltage to safer levels for distribution and use in homes and industries.

55. The property of electric current which is applicable in the fuse wire

The property of electric current which is applicable in the fuse wire is

[amp_mcq option1=”chemical effect of current” option2=”magnetic effect of current” option3=”heating effect of current” option4=”optical property of current” correct=”option3″]

This question was previously asked in
UPSC NDA-2 – 2016
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The correct answer is C) heating effect of current. A fuse wire is designed to melt and break the circuit when the current flowing through it exceeds a certain limit. This melting is caused by the heat generated due to the resistance of the wire when current passes through it, which is known as the heating effect of electric current (Joule’s Law).
– Fuse wire protects electrical circuits from overcurrents.
– Its operation relies on melting due to excessive heat.
– The heat generated is proportional to the square of the current and the resistance of the wire (H = I²Rt).
– Chemical effect of current is related to electrolysis.
– Magnetic effect of current is used in devices like motors, relays, and electromagnets.
– Optical property of current is not a standard term; effects like light emission (LEDs) are related to current flow but not the principle behind a fuse.

56. When three resistors, each having resistance r, are connected in paral

When three resistors, each having resistance r, are connected in parallel, their resultant resistance is x. If these three resistances are connected in series, the total resistance will be

[amp_mcq option1=”3x” option2=”3rx” option3=”9x” option4=”3/x” correct=”option3″]

This question was previously asked in
UPSC NDA-2 – 2016
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When three resistors, each with resistance $r$, are connected in parallel, their resultant resistance $x$ is given by:
$\frac{1}{x} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{3}{r}$
So, $x = \frac{r}{3}$, which means $r = 3x$.
When these three resistors are connected in series, the total resistance is the sum of individual resistances:
Total resistance $R_{series} = r + r + r = 3r$.
Substituting the expression for $r$ in terms of $x$ ($r=3x$):
$R_{series} = 3 \times (3x) = 9x$.
Resistors in parallel: $\frac{1}{R_{parallel}} = \sum \frac{1}{R_i}$. Resistors in series: $R_{series} = \sum R_i$.
Parallel connections decrease the total resistance, while series connections increase the total resistance. For identical resistors, the parallel resistance is $R/n$ and the series resistance is $nR$, where $R$ is the individual resistance and $n$ is the number of resistors.

57. Which of the following items is used in the household wirings to preve

Which of the following items is used in the household wirings to prevent accidental fire in case of short circuit ?

[amp_mcq option1=”Insulated wire” option2=”Plastic switches” option3=”Non-metallic coatings on conducting wires” option4=”Electric fuse” correct=”option4″]

This question was previously asked in
UPSC NDA-2 – 2016
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The correct answer is D. An electric fuse is a safety device specifically designed to protect electrical circuits and prevent fires caused by overcurrents, such as those resulting from a short circuit.
In case of a short circuit or overload, an excessively large current flows through the circuit. This large current causes the fuse wire, which has a specific melting point and resistance, to heat up rapidly and melt. When the fuse wire melts, it breaks the circuit, interrupting the flow of current and preventing overheating of wires and appliances, thus preventing accidental fire.
Insulated wires (A) and non-metallic coatings (C) are crucial for preventing *initial* contact between conductors that could cause a short circuit, but they do not protect against the consequences *after* a short circuit or overload has occurred. Plastic switches (B) are used for safely opening and closing circuits under normal operating conditions, not for overcurrent protection. Modern installations often use Miniature Circuit Breakers (MCBs) or Residual Current Devices (RCDs) in addition to or instead of traditional fuses for overcurrent and fault protection.

58. The resistance of a wire of length / and area of cross-section a is x

The resistance of a wire of length / and area of cross-section a is x ohm. If the wire is stretched to double its length, its resistance would become:

[amp_mcq option1=”2x ohm” option2=”0.5 x ohm” option3=”4x ohm” option4=”6x ohm” correct=”option3″]

This question was previously asked in
UPSC NDA-2 – 2015
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The resistance (R) of a wire is given by R = ρ * (l/a), where ρ is resistivity, l is length, and a is cross-sectional area. If the wire is stretched to double its length (l’ = 2l), its volume (V = a * l) remains constant. So, a’ * l’ = a * l => a’ * (2l) = a * l => a’ = a/2. The new resistance is R’ = ρ * (l’/a’) = ρ * (2l / (a/2)) = ρ * (4l/a) = 4 * (ρ * l/a) = 4x.
When a wire is stretched, its length increases, and its cross-sectional area decreases while its volume remains constant. This significantly increases its resistance.
The resistance is directly proportional to length and inversely proportional to cross-sectional area. Stretching the wire affects both parameters simultaneously.

59. Three equal resistances when combined in series are equivalent to 90 o

Three equal resistances when combined in series are equivalent to 90 ohm. Their equivalent resistance when combined in parallel will be:

[amp_mcq option1=”10 ohm” option2=”30 ohm” option3=”270 ohm” option4=”810 ohm” correct=”option1″]

This question was previously asked in
UPSC NDA-2 – 2015
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Let R be the resistance of each of the three equal resistors. When connected in series, the equivalent resistance (Rs) is the sum of individual resistances: Rs = R + R + R = 3R. Given Rs = 90 ohm, we have 3R = 90 ohm, so R = 30 ohm. When the three equal resistors are connected in parallel, the equivalent resistance (Rp) is given by the formula 1/Rp = 1/R + 1/R + 1/R = 3/R. Thus, Rp = R/3. Substituting R = 30 ohm, we get Rp = 30 ohm / 3 = 10 ohm.
Calculation of equivalent resistance for resistors connected in series and parallel.
For N equal resistors of resistance R, the equivalent resistance in series is NR, and the equivalent resistance in parallel is R/N.

60. Lightning conductors are used to protect building from lightning strik

Lightning conductors are used to protect building from lightning strikes. Which of the following statements is / are true about lightning conductors?

  • 1. Lightning conductors create an electric field at its top so that lightning strikes it preferentially
  • 2. Lightning conductors reduce the effect of the strike by uniformly distributing the charge (current) over the surface of the building
  • 3. Lightning conductors take all charge (current) to deep down in the earth
  • 4. Lightning conductors must be installed at a place taller than the building

Select the correct answer using the code given below:

[amp_mcq option1=”1 and 2″ option2=”3 and 4 only” option3=”1, 3 and 4″ option4=”4 only” correct=”option3″]

This question was previously asked in
UPSC NDA-2 – 2015
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The correct answer is C. Statements 1, 3, and 4 are true about lightning conductors.
– Statement 1 is correct. The pointed tip of a lightning conductor concentrates electric charge, creating a strong electric field. This strong field can ionize the air, making the conductor the preferred point of strike if lightning occurs nearby.
– Statement 2 is incorrect. Lightning conductors provide a direct, low-resistance path to the earth, diverting the lightning current away from the building’s structure, not distributing it over the surface.
– Statement 3 is correct. The conductor system is connected to grounding rods or plates buried deep in the earth, allowing the immense electrical charge from a lightning strike to dissipate safely into the ground.
– Statement 4 is correct. For a lightning conductor to effectively intercept a strike, it must be the highest point of the structure it is protecting.
Lightning conductors, invented by Benjamin Franklin, protect buildings by safely channeling the electrical discharge from a lightning strike to the ground, preventing damage from fire or structural breakdown caused by the current passing through the building itself.
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