Electric Current

31. An electric bulb is connected to 220 V generator. The current drawn is

An electric bulb is connected to 220 V generator. The current drawn is 600 mA. What is the power of the bulb?

132 W
13.2 W
1320 W
13200 W
This question was previously asked in
UPSC NDA-2 – 2022
Download PDFAttempt Online
The power (P) of an electrical device connected to a circuit is given by the formula:
P = V × I
Where V is the voltage across the device and I is the current flowing through it.
Given:
Voltage (V) = 220 V
Current (I) = 600 mA
First, convert the current from milliamperes (mA) to amperes (A):
1 A = 1000 mA
So, 600 mA = 600 / 1000 A = 0.6 A.
Now, calculate the power:
P = 220 V × 0.6 A
P = 220 × (6/10) W
P = 22 × 6 W
P = 132 W.
Power is calculated as the product of voltage and current (P=VI). Ensure units are in Volts and Amperes to get Power in Watts.
This calculation is a fundamental application of Ohm’s law and the power formula in basic electricity. The resistance of the bulb could also be calculated using Ohm’s law (V=IR), but it is not needed to find the power. The power rating (in Watts) indicates the rate at which the bulb consumes electrical energy.

32. Which one of the following statements regarding a current-carrying sol

Which one of the following statements regarding a current-carrying solenoid is not correct?

The magnetic field inside the solenoid is uniform.
The current-carrying solenoid behaves like a bar magnet.
The magnetic field inside the solenoid increases with increase in current.
If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.
This question was previously asked in
UPSC NDA-2 – 2022
Download PDFAttempt Online
Let’s analyze each statement about a current-carrying solenoid:
A) The magnetic field inside a *long* solenoid is approximately uniform and directed along the axis of the solenoid, except near the ends. This statement is correct for an ideal or long solenoid.
B) A current-carrying solenoid creates a magnetic field pattern similar to that of a bar magnet, with magnetic poles at its ends. This statement is correct.
C) The magnitude of the magnetic field inside a solenoid is given by $B = \mu n I$, where $\mu$ is the permeability of the core material, $n$ is the number of turns per unit length, and $I$ is the current. The field is directly proportional to the current ($I$). So, increasing the current increases the magnetic field. This statement is correct.
D) If a soft iron bar (a ferromagnetic material with high permeability) is inserted inside the solenoid, the magnetic field inside increases significantly. This happens because the soft iron gets strongly magnetized in the direction of the solenoid’s field, and its own magnetic field adds to the field produced by the current. The permeability of soft iron is much greater than the permeability of air or vacuum ($\mu >> \mu_0$). The magnetic field does *not* remain the same; it increases. This statement is incorrect.
Ferromagnetic materials like soft iron dramatically increase the magnetic field strength when placed inside a solenoid or coil because they become strongly magnetized, effectively increasing the magnetic permeability of the core.
This property is used to make electromagnets, where a soft iron core is inserted into a solenoid to produce a very strong magnetic field when current flows.

33. The commercial unit of electrical energy is kilowatt-hour (kWh), which

The commercial unit of electrical energy is kilowatt-hour (kWh), which is equal to

$3.6 imes 10^6$ J
$3.6 imes 10^3$ J
$10^3$ J
1 J
This question was previously asked in
UPSC NDA-2 – 2022
Download PDFAttempt Online
The commercial unit of electrical energy is kilowatt-hour (kWh). Energy consumed is calculated as Power multiplied by Time.
1 kilowatt (kW) = 1000 Watts (W)
1 hour (h) = 3600 seconds (s)
Energy (Joules) = Power (Watts) × Time (seconds)
1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3,600,000 Joules (J)
In scientific notation, this is $3.6 \times 10^6$ J.
1 kWh is the energy used by a 1 kW device operating for 1 hour. The conversion factor from kilowatt-hour to Joules is obtained by converting kilowatts to watts and hours to seconds.
The Joule is the SI unit of energy, but for practical purposes, especially for calculating electricity consumption in households and industries, kilowatt-hour is commonly used. It represents a larger unit of energy. 1 unit of electricity on a meter typically corresponds to 1 kWh.

34. The device used to produce electric current is known as

The device used to produce electric current is known as

motor
generator
ammeter
galvanometer
This question was previously asked in
UPSC NDA-2 – 2021
Download PDFAttempt Online
The correct answer is B) generator.
A generator is a device that converts mechanical energy into electrical energy, thereby producing electric current.
A motor converts electrical energy into mechanical energy. An ammeter is used to measure electric current. A galvanometer is used to detect or measure small electric currents.

35. Imagine a current-carrying straight conductor with magnetic field of l

Imagine a current-carrying straight conductor with magnetic field of lines in anti-clockwise direction. Then the direction of current is determined by

the Right-Hand Thumb rule and it would be in the downward direction.
the Left-Hand Thumb rule and it would be in the downward direction.
the Right-Hand Thumb rule and it would be in the upward direction.
the Left-Hand Thumb rule and it would be in the upward direction.
This question was previously asked in
UPSC NDA-2 – 2021
Download PDFAttempt Online
The direction of the magnetic field around a straight current-carrying conductor is determined by the Right-Hand Thumb Rule (also known as Ampere’s Right-Hand Rule). According to this rule, if you point the thumb of your right hand in the direction of the conventional current, your fingers curl in the direction of the magnetic field lines around the wire. The question states the magnetic field lines are in an anti-clockwise direction when viewed from a certain perspective (presumably from above the conductor). If the current were downwards, using the right-hand rule, your fingers would curl clockwise. If the current were upwards, your fingers would curl anti-clockwise. Since the magnetic field is anti-clockwise, the current must be in the upward direction, and the rule used is the Right-Hand Thumb rule.
The Right-Hand Thumb Rule for a straight conductor states: Point your right thumb in the direction of the conventional current. Your fingers curl in the direction of the magnetic field. An anti-clockwise magnetic field (when looking down onto the wire) implies the current is flowing upwards.
The Left-Hand Rule is typically used in contexts involving forces on current-carrying conductors in magnetic fields (Fleming’s Left-Hand Rule) or for the direction of force on a moving charge in a magnetic field. For determining the direction of the magnetic field created by a current, the Right-Hand Rule is used.

36. Three equal resistors are connected in parallel configuration in a clo

Three equal resistors are connected in parallel configuration in a closed electrical circuit. Then the total resistance in the circuit becomes

one-third of the individual resistance.
two-third of the individual resistance.
equal to the individual resistance.
three times of the individual resistance.
This question was previously asked in
UPSC NDA-2 – 2021
Download PDFAttempt Online
When resistors are connected in parallel, the total resistance is calculated using the formula for parallel resistances. If three equal resistors, each with resistance R, are connected in parallel, the reciprocal of the total resistance ($R_{total}$) is the sum of the reciprocals of the individual resistances: $1/R_{total} = 1/R + 1/R + 1/R = 3/R$. Therefore, $R_{total} = R/3$. The total resistance becomes one-third of the individual resistance.
For ‘n’ equal resistors each of resistance ‘R’ connected in parallel, the total resistance is $R_{total} = R/n$. In this case, n=3, so $R_{total} = R/3$.
In contrast, if the resistors were connected in series, the total resistance would be the sum of the individual resistances: $R_{total} = R + R + R = 3R$. Connecting resistors in parallel decreases the total resistance of the circuit, while connecting them in series increases it.

37. When the short circuit condition occurs, the current in the circuit

When the short circuit condition occurs, the current in the circuit

becomes zero
remains constant
increases substantially
keeps on changing randomly
This question was previously asked in
UPSC NDA-2 – 2020
Download PDFAttempt Online
A short circuit condition occurs when a low-resistance path is created between two points in a circuit that normally have a potential difference across them. According to Ohm’s Law (I = V/R), if the voltage (V) source maintains a constant potential difference and the resistance (R) in the circuit drops significantly (approaching zero in an ideal short circuit), the current (I) will increase substantially. In a real circuit, the current increase is limited by the internal resistance of the power source and the resistance of the wires themselves, but it still becomes much higher than the normal operating current, often leading to overheating and damage.
In a short circuit, resistance becomes very low, leading to a very high current according to Ohm’s Law (I = V/R).
Short circuits can be dangerous as they can cause wires to melt, fires, or damage to electrical equipment. Circuit breakers or fuses are designed to detect and interrupt the circuit during a short circuit condition to prevent such hazards.

38. Consider the following image: A proton enters a magnetic field at righ

Consider the following image:
A proton enters a magnetic field at right angles to it, as shown above. The direction of force acting on the proton will be

to the right
to the left
out of the page
into the page
This question was previously asked in
UPSC NDA-2 – 2020
Download PDFAttempt Online
The force acting on a charged particle in a magnetic field is given by the Lorentz force formula, F = q(v x B), where q is the charge, v is the velocity, and B is the magnetic field. For a positive charge like a proton (q > 0), the direction of the force is the same as the direction of the vector cross product v x B. The diagram shows the velocity (v) of the proton pointing to the right and the magnetic field (B) represented by crosses (X), which conventionally indicates a vector pointing into the plane of the page. Using the Right Hand Rule for the cross product v x B: Point the fingers of your right hand in the direction of v (right). Curl your fingers towards the direction of B (into the page). Your thumb points in the direction of v x B, which is upwards (in the plane of the page). Since the proton is positively charged, the force is also upwards. However, ‘upwards’ is not provided as an option. This suggests a potential issue with the question or options. Assuming there is an intended correct answer among the options, the most plausible alternative interpretation that leads to a provided option is if the crosses represent a magnetic field pointing downwards (in the plane of the page) while the velocity is right (in the plane of the page). In this case, F = q(v x B) where v is right (+x) and B is downwards (-y). v x B = (+x) x (-y) = -z, which corresponds to ‘into the page’ (assuming +z is out of the page). Thus, under this alternative interpretation (where crosses represent a downwards field), the force is into the page (Option D). This explanation is based on the likelihood of an intended answer despite the contradiction with standard diagram conventions.
The direction of the Lorentz force F on a positive charge is given by the Right Hand Rule applied to the cross product v x B. Standard convention for crosses (X) is ‘into the plane of the diagram’.
Applying standard physics principles to the diagram (v right, B into page) correctly yields a force direction upwards. Since this is not an option, the question as presented is likely flawed. However, in test scenarios, one might need to infer the most probable intended scenario that matches an option.

39. A metallic wire having resistance of 20 Ω is cut into two equal parts

A metallic wire having resistance of 20 Ω is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to

20 Ω
10 Ω
5 Ω
15 Ω
This question was previously asked in
UPSC NDA-2 – 2020
Download PDFAttempt Online
The original metallic wire has a resistance R = 20 Ω. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R₁ = R₂ = R/2 = 20 Ω / 2 = 10 Ω. When these two parts are connected in parallel, the equivalent resistance (Rₚ) is given by the formula 1/Rₚ = 1/R₁ + 1/R₂. Substituting the values, 1/Rₚ = 1/10 Ω + 1/10 Ω = 2/10 Ω = 1/5 Ω. Therefore, the equivalent resistance Rₚ = 5 Ω.
The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.
For two resistors R₁ and R₂ in parallel, the equivalent resistance can also be calculated as Rₚ = (R₁ * R₂) / (R₁ + R₂). In this case, Rₚ = (10 Ω * 10 Ω) / (10 Ω + 10 Ω) = 100 / 20 = 5 Ω.

40. Two equal resistors R are connected in parallel, and a battery of 12 V

Two equal resistors R are connected in parallel, and a battery of 12 V is connected across this combination. A dc current of 100 mA flows through the circuit as shown below :
The value of R is

120 Ω
240 Ω
60 Ω
100 Ω
This question was previously asked in
UPSC NDA-2 – 2020
Download PDFAttempt Online
Two equal resistors R are connected in parallel. The equivalent resistance (Req) of two equal resistors in parallel is given by Req = R/2. According to Ohm’s Law, the voltage (V) across the circuit is equal to the current (I) flowing through it multiplied by the equivalent resistance (Req): V = I * Req.
Given V = 12 V and I = 100 mA = 0.1 A.
So, 12 V = 0.1 A * (R/2).
Rearranging the equation to solve for R:
12 = 0.1 * R / 2
12 = 0.05 * R
R = 12 / 0.05
R = 12 / (5/100)
R = 12 * (100/5)
R = 12 * 20
R = 240 Ω.
For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: 1/Req = 1/R1 + 1/R2 + … For two equal resistors R, 1/Req = 1/R + 1/R = 2/R, so Req = R/2.
Ohm’s Law (V = IR) is fundamental to analyzing simple electric circuits. Units must be consistent (Volts, Amperes, Ohms). Milliamperes (mA) must be converted to Amperes (A) by dividing by 1000 (100 mA = 0.1 A).
Page 1Page 2Page 3Page 4Page 5
Exit mobile version