Electric Current Archives

21. For an electric circuit given below, the correct combination of voltag

For an electric circuit given below, the correct combination of voltage (V) and current (I) is

V = 900 V; I = 18 A

V = 300 V; I = 5.5 A

V = 600 V; I = 1 A

V = 300 V; I = 2 A

This question was previously asked in

UPSC NDA-2 – 2024

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Assuming the circuit diagram (not visible) represents a setup where a voltage of 300 V results in a current of 2 A, then option D is the correct combination. This implies the equivalent resistance of the circuit connected to the voltage source is 150 Ω (since V = IR, R = V/I = 300V / 2A = 150 Ω).

– Ohm’s Law (V = IR) relates the voltage (V) across a component or circuit, the current (I) flowing through it, and its resistance (R).
– For a complete circuit with a voltage source and an equivalent resistance R_eq, the total voltage V supplied by the source is related to the total current I flowing out of the source by V = I * R_eq.
– The question provides possible pairs of total voltage (V) and total current (I) for an electric circuit. We can calculate the implied equivalent resistance (R_eq = V/I) for each pair:
– A) V = 900 V, I = 18 A => R_eq = 900/18 = 50 Ω
– B) V = 300 V, I = 5.5 A => R_eq = 300/5.5 ≈ 54.55 Ω
– C) V = 600 V, I = 1 A => R_eq = 600/1 = 600 Ω
– D) V = 300 V, I = 2 A => R_eq = 300/2 = 150 Ω
– The circuit diagram would define the value of R_eq based on the configuration and values of its components (resistors, etc.). Without the diagram, we must assume that one of these calculated R_eq values corresponds to the actual circuit’s equivalent resistance. Assuming option D is correct, the circuit must have an equivalent resistance of 150 Ω when connected to a 300 V source or when a 2 A current flows through it resulting in a 300 V drop.

To definitively solve this problem, the circuit diagram showing the voltage source(s) and resistor network is required. The calculation of the equivalent resistance from the diagram would then confirm which (V, I) pair from the options is consistent with Ohm’s Law and the circuit’s resistance. Standard resistor values and combinations often yield resistances like 50, 150, 600 ohms.

22. In which one among the following situations, the bulb would glow the m

In which one among the following situations, the bulb would glow the most ? (Consider all batteries are the same)

Diagram (a)

Diagram (b)

Diagram (c)

Diagram (d)

This question was previously asked in

UPSC NDA-2 – 2024

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Assuming Diagram (b) shows two identical batteries connected in series to a single bulb, this configuration would result in the highest voltage across the bulb, making it glow the brightest.

– The brightness of a bulb (assumed to be resistive) is proportional to the power dissipated, P. Power can be calculated as P = V²/R or P = I²R, where V is the voltage across the bulb, I is the current through the bulb, and R is its resistance. For a given bulb (fixed R), maximum brightness corresponds to maximum voltage or maximum current.
– Assuming each battery provides a voltage V:
– Diagram (a) (one battery, one bulb): Voltage across bulb = V.
– Diagram (b) (two batteries in series, one bulb): The voltages of batteries in series add up. Total voltage = 2V. This voltage is across the bulb.
– Diagram (c) (two batteries in parallel, one bulb): Batteries in parallel maintain the same voltage as a single battery (assuming identical ideal batteries). Total voltage = V. This voltage is across the bulb. (Parallel connection increases total current capacity, but not voltage).
– Diagram (d) (one battery, two bulbs in series): The voltage V from the battery is divided between the two bulbs. Assuming identical bulbs, voltage across each bulb = V/2.
– Comparing the voltage across the bulb in each case: (a) V, (b) 2V, (c) V, (d) V/2.
– Since the power dissipated (and thus brightness) is proportional to V², the bulb will glow brightest when the voltage across it is highest, which is in case (b) with two batteries in series (2V).

Connecting batteries in series increases the total voltage supplied, while connecting identical batteries in parallel increases the total charge capacity and allows the supply of a larger current for a longer duration, without increasing the voltage (ideally). Connecting bulbs in series increases the total resistance of the circuit, reducing the current and the voltage across each individual bulb (and thus brightness, assuming they are identical).

23. Lightning is due to

Lightning is due to

The flow of charges between different parts of the cloud

The short-circuiting of charges between the upper and lower surfaces of the cloud

The collection of positively charged particles on the base and collection of negatively charged particles at the top of the cloud

The induction of positive charge on the ground below the negative charge at the base of the cloud

This question was previously asked in

UPSC NDA-2 – 2024

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The correct answer is B. Lightning is fundamentally a rapid, large-scale electrical discharge, often described as a “short-circuiting” of charges, which frequently occurs between different charged regions within a thundercloud, typically between the upper and lower parts.

– Lightning is an electrical discharge phenomena occurring in the atmosphere.
– It is caused by the build-up of electric charge differences within cumulonimbus clouds (thunderclouds) or between the cloud and the ground.
– Charge separation within a cloud typically results in positive charges accumulating at the top and negative charges accumulating at the bottom, although complex charge distributions exist.
– When the potential difference between these regions (or between the cloud and ground) exceeds the dielectric strength of the air, an electrical breakdown occurs, creating a conductive channel.
– The lightning strike is the rapid flow of electric charge (electrons and ions) through this channel. This discharge can happen within the cloud (intra-cloud lightning), between clouds (inter-cloud lightning), or between the cloud and the ground (cloud-to-ground lightning).
– Option B describes the rapid discharge (“short-circuiting”) between upper and lower charged regions within a cloud, which is a common type of lightning (intra-cloud lightning). While option A is also a flow of charges, B is more specific about the location and uses the term “short-circuiting” which captures the rapid discharge aspect. Option C describes the cause (charge separation) but gets the common polarity wrong (usually positive at top, negative at bottom). Option D describes ground charge induction, a condition for ground strikes, not the general cause of lightning.

Cloud-to-ground lightning is the most dangerous type. The rapid heating of the air along the lightning channel causes it to expand explosively, creating the sound waves we perceive as thunder.

24. Two resistances of 5.0 Ω and 7.0 Ω are connected in series and the com

Two resistances of 5.0 Ω and 7.0 Ω are connected in series and the combi- nation is connected in parallel with a resistance of 36.0 Ω. The equivalent resistance of the combination of three resistors is

24.0 Ω

12.0 Ω

9.0 Ω

6.0 Ω

This question was previously asked in

UPSC NDA-2 – 2024

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The correct answer is C) 9.0 Ω.

First, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances:
R_series = R₁ + R₂ = 5.0 Ω + 7.0 Ω = 12.0 Ω.
Next, this series combination (with R_series = 12.0 Ω) is connected in parallel with a third resistance (R₃ = 36.0 Ω). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances:
1/R_eq = 1/R_series + 1/R₃
1/R_eq = 1/12.0 Ω + 1/36.0 Ω
To add the fractions, find a common denominator, which is 36.
1/R_eq = (3/36) + (1/36) = 4/36
1/R_eq = 1/9
R_eq = 9.0 Ω.

Understanding how to combine resistances in series and parallel is fundamental in circuit analysis. Resistances in series add directly, increasing the total resistance. Resistances in parallel combine in a way that the reciprocal of the total resistance is the sum of the reciprocals, resulting in a lower total resistance than the smallest individual resistance.

25. What will happen if a collection of positive and negative charges are

What will happen if a collection of positive and negative charges are passed at a high speed through a magnetic field which is perpendicular to the direction of motion of the charges ? (Assume that both kind of charges are NOT going to recombine)

Both kind of charges will stop moving

Positive charges and negative charges will separate out

Positive charges will stop but negative charges will continue moving uninterrupted

Both kind of charges will keep moving uninterrupted

This question was previously asked in

UPSC NDA-2 – 2023

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When a collection of positive and negative charges are passed at high speed through a magnetic field perpendicular to their motion, the positive and negative charges will separate out.

The Lorentz force ($\vec{F} = q(\vec{v} \times \vec{B})$) acts on a moving charge in a magnetic field. The force is perpendicular to both the velocity ($\vec{v}$) and the magnetic field ($\vec{B}$). For positive charges, the force direction is given by the right-hand rule applied to $\vec{v} \times \vec{B}$. For negative charges (where $q$ is negative), the force is in the opposite direction. This difference in the direction of the force causes the positive and negative charges to be deflected in opposite directions, leading to separation.

This principle is utilized in devices like mass spectrometers or velocity selectors to separate charged particles based on their mass-to-charge ratio or velocity. The charges will move in curved paths, not stop or continue uninterrupted, as long as they are within the magnetic field.

26. Ms. Rani decides to convert her AC generator into a DC generator. Whic

Ms. Rani decides to convert her AC generator into a DC generator. Which one of the following she would need to use?

A split-ring type commutator

Slip rings and brushes

A stronger magnetic field

A rectangular wire loop

This question was previously asked in

UPSC NDA-2 – 2023

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A) A split-ring type commutator is the correct component needed to convert an AC generator into a DC generator.

– An AC generator produces alternating current due to the use of slip rings, which maintain continuous contact with the rotating coil.
– To convert this to direct current (DC), the direction of the current in the external circuit needs to be maintained constant.
– A split-ring commutator achieves this by reversing the connection of the coil to the external circuit every half rotation, ensuring that the current flows in the same direction in the external load.

– Slip rings are used in AC generators to connect the armature coil to the external circuit.
– A stronger magnetic field would increase the magnitude of the induced electromotive force (EMF) and current, but would not change the nature of the output from AC to DC.
– A rectangular wire loop is the armature coil, which is a component of both AC and DC generators. The type of output depends on how the coil is connected to the external circuit (via slip rings for AC, or a commutator for DC).

27. The potential difference between the two end terminals of an electric

The potential difference between the two end terminals of an electric heater is 220 V and the current through it is 0·5 A. What would be the current through the heater if the potential difference across the terminals of the heater is reduced to 120 V ?

1·0 A

0·5 A

0·27 A

0·7 A

This question was previously asked in

UPSC NDA-2 – 2023

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Assuming the resistance of the electric heater remains constant, the current through it would be approximately 0.27 A when the potential difference is reduced to 120 V.

– We can use Ohm’s Law, which states that the voltage (V) across a resistor is directly proportional to the current (I) flowing through it, provided the temperature and other physical conditions remain unchanged (V = I * R).
– First, calculate the resistance (R) of the heater using the initial conditions: V1 = 220 V, I1 = 0.5 A.
– R = V1 / I1 = 220 V / 0.5 A = 440 ohms.
– Now, use this resistance to find the new current (I2) when the potential difference (V2) is 120 V.
– V2 = I2 * R
– 120 V = I2 * 440 ohms
– I2 = 120 V / 440 ohms = 12 / 44 A = 3 / 11 A.
– Calculating the decimal value: 3 / 11 ≈ 0.2727 A.

In reality, the resistance of a heater element made of materials like nichrome increases with temperature. However, standard Ohm’s Law problems involving fixed resistors or devices like heaters typically assume constant resistance unless otherwise specified. The calculated current represents the value based on this common assumption.

28. Which one of the following terms cannot represent electrical power in

Which one of the following terms cannot represent electrical power in a circuit?

I^2/R

I^2R

V^2/R

This question was previously asked in

UPSC NDA-2 – 2022

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Electrical power (P) in a circuit can be calculated using different formulas derived from the relationship between voltage (V), current (I), and resistance (R), as defined by Ohm’s Law (V = IR). The standard formulas for power are:
1. P = VI (Power = Voltage × Current)
2. P = I²R (Substituting V = IR into P = VI: P = (IR)I = I²R)
3. P = V²/R (Substituting I = V/R into P = VI: P = V(V/R) = V²/R)
Looking at the options:
A) VI – This is a correct formula for power.
B) I²/R – This is incorrect. The correct formula involving I and R is I²R.
C) I²R – This is a correct formula for power.
D) V²/R – This is a correct formula for power.

Understanding the relationships between power, voltage, current, and resistance through Ohm’s Law is fundamental to electrical circuit analysis.

The power dissipated by a resistor represents the rate at which electrical energy is converted into heat energy. These power formulas are applicable to resistive components in both DC and AC circuits (though for AC, power can also involve a power factor).

29. A DC generator works on the principle of

A DC generator works on the principle of

Ohm's law

Joule's law of heating

Faraday's laws of electromagnetic induction

None of the above

This question was previously asked in

UPSC NDA-2 – 2022

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A DC generator works on the principle of Faraday’s laws of electromagnetic induction.

Electromagnetic induction is the process where a conductor moving through a magnetic field (or a changing magnetic field near a conductor) induces an electromotive force (EMF), which can drive an electric current. Faraday’s laws quantify this phenomenon.

A generator utilizes mechanical energy (to rotate a coil or magnetic field) to cause a change in magnetic flux through a coil, thereby inducing an electric current according to Faraday’s laws. Ohm’s law relates voltage, current, and resistance in a circuit. Joule’s law of heating describes the heat produced by current flowing through a resistor. A motor works on the principle of the force experienced by a current-carrying conductor in a magnetic field.

30. A current of 0·6 A is drawn by an electric bulb for 10 minutes. Which

A current of 0·6 A is drawn by an electric bulb for 10 minutes. Which one of the following is the amount of electric charge that flows through the circuit?

6 C

0·6 C

360 C

36 C

This question was previously asked in

UPSC NDA-2 – 2022

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The amount of electric charge that flows through the circuit is 360 C.

Electric charge (Q) is defined as the product of electric current (I) and time (t): Q = I × t.
The current given is I = 0.6 A.
The time given is t = 10 minutes. This needs to be converted to seconds: 10 minutes × 60 seconds/minute = 600 seconds.

Now, calculate the charge: Q = 0.6 A × 600 s = 360 Ampere-seconds. The unit of electric charge is the Coulomb (C), where 1 Coulomb = 1 Ampere × 1 second. Therefore, Q = 360 C.