Electric Current

11. Let an electric current of 1.5 A flow through an incandescent lamp in

Let an electric current of 1.5 A flow through an incandescent lamp in a circuit. What is the amount of charge that flows through it in 10 ms?

[amp_mcq option1=”0.015 C” option2=”0.15 C” option3=”1.5 C” option4=”15 C” correct=”option1″]

This question was previously asked in
UPSC CAPF – 2021
Download PDFAttempt Online
Electric current (I) is defined as the rate of flow of electric charge (Q) per unit time (t). The relationship is given by Q = I * t.
Given the current (I) = 1.5 A and the time (t) = 10 ms. First, convert the time into seconds: 10 ms = 10 * 10^-3 seconds = 0.01 seconds. Now, calculate the charge: Q = 1.5 A * 0.01 s = 0.015 Coulombs (C).
The unit of electric current is the Ampere (A), and the unit of electric charge is the Coulomb (C). One Ampere is equal to the flow of one Coulomb of charge per second (1 A = 1 C/s). Milliseconds (ms) are a common unit of time, where 1 second = 1000 milliseconds.

12. Which one of the following statements regarding Fleming’s Rule is co

Which one of the following statements regarding Fleming’s Rule is correct?

[amp_mcq option1=”Fleming’s left hand rule gives direction of force on a current-carrying conductor in a magnetic field” option2=”Fleming’s right hand rule gives direction of force on a current-carrying conductor in a magnetic field” option3=”Both the left-hand and right-hand rule can be used for finding direction of force on a current-carrying conductor in a magnetic field” option4=”Fleming’s rules has nothing to do with magnetic field” correct=”option1″]

This question was previously asked in
UPSC CAPF – 2020
Download PDFAttempt Online
Fleming’s left hand rule gives direction of force on a current-carrying conductor in a magnetic field.
Fleming’s Left-Hand Rule is used to determine the direction of the force (thrust or motion) on a current-carrying conductor when placed in an external magnetic field. It relates the directions of the magnetic field, the electric current, and the resulting force. The thumb represents the direction of the force, the forefinger represents the direction of the magnetic field, and the middle finger represents the direction of the current.
Fleming’s Right-Hand Rule is used to determine the direction of induced current when a conductor moves in a magnetic field. It is used in generators. The Left-Hand Rule is primarily used in motors.

13. Two magnetic field lines produced by the same source

Two magnetic field lines produced by the same source

[amp_mcq option1=”never intersect” option2=”can originate from same point” option3=”can terminate at same point” option4=”can intersect depending on situation” correct=”option1″]

This question was previously asked in
UPSC CAPF – 2020
Download PDFAttempt Online
Two magnetic field lines produced by the same source never intersect.
Magnetic field lines represent the direction of the magnetic field at different points. If two magnetic field lines were to intersect at a point, it would mean that the magnetic field at that point has two different directions simultaneously, which is impossible.
Magnetic field lines are continuous curves forming closed loops (outside the magnet from North to South, inside from South to North). They are denser where the magnetic field is stronger. They originate from the North pole and terminate at the South pole outside the magnet.

14. Directions : The following eight (8) items consist of two statements,

Directions :
The following eight (8) items consist of two statements, Statement I and Statement II. Examine these two statements carefully and select the correct answer using the code given below.
Code :
Statement I : A compass needle placed near a current-carrying wire will get deflected.
Statement II : A current-carrying wire creates magnetic field around it.

[amp_mcq option1=”Both the statements are individually true and Statement II is the correct explanation of Statement I” option2=”Both the statements are individually true but Statement II is not the correct explanation of Statement I” option3=”Statement I is true but Statement II is false” option4=”Statement I is false but Statement II is true” correct=”option1″]

This question was previously asked in
UPSC CAPF – 2018
Download PDFAttempt Online
The correct answer is A) Both the statements are individually true and Statement II is the correct explanation of Statement I.
– Statement I: A compass needle is a small magnet. It aligns itself with the local magnetic field. If a compass needle is placed near a current-carrying wire, it will experience a force due to the magnetic field produced by the current, causing it to deflect. This statement is true.
– Statement II: A current-carrying wire produces a magnetic field around it. This phenomenon was discovered by Hans Christian Ørsted and is a fundamental principle of electromagnetism, described by laws like Ampère’s law. This statement is true.
– Statement II provides the reason why the compass needle in Statement I gets deflected. The magnetic field created by the current-carrying wire (Statement II) is what causes the deflection of the compass needle (Statement I).
The direction and strength of the magnetic field around a straight current-carrying wire can be determined using the right-hand rule and the Biot-Savart Law or Ampère’s Law. The deflection of the compass needle demonstrates the presence and direction of this magnetic field.

15. Let a resistor having 4 ohm resistance be connected across the termina

Let a resistor having 4 ohm resistance be connected across the terminals of a 12 volt battery. Then the charge in coulomb passing through the resistor per second is:

[amp_mcq option1=”12″ option2=”4″ option3=”3″ option4=”0.33″ correct=”option3″]

This question was previously asked in
UPSC CAPF – 2016
Download PDFAttempt Online
The correct option is C) 3.
According to Ohm’s Law, the current (I) flowing through a resistor is directly proportional to the voltage (V) across it and inversely proportional to its resistance (R). The relationship is given by the formula V = IR. The current (I) is also defined as the amount of electric charge (Q) passing through a point per unit time (t), i.e., I = Q/t. The question asks for the charge passing through the resistor *per second*, which means we need to find Q when t = 1 second. From I = Q/t, we get Q = I * t. If t = 1 second, then Q = I * 1 = I. Therefore, the charge passing through the resistor per second is equal to the current flowing through it.
Given: Voltage (V) = 12 volts, Resistance (R) = 4 ohm.
Using Ohm’s Law, I = V / R = 12 V / 4 ohm = 3 Amperes.
The current is 3 Amperes, which means 3 Coulombs of charge flow through the resistor every second.
Charge passing per second = Current (I) = 3 Coulombs.

16. Transformer is used to 1. convert low a.c. voltage to high voltage

Transformer is used to

  • 1. convert low a.c. voltage to high voltage
  • 2. convert high a.c. voltage to low voltage
  • 3. convert direct current to alternating current
  • 4. regulate the fluctuation of voltage

Select the correct answer using the code given below.

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

This question was previously asked in
UPSC CAPF – 2013
Download PDFAttempt Online
The correct option is C. A transformer is used to convert low a.c. voltage to high voltage and convert high a.c. voltage to low voltage.
Transformers are electrical devices that transfer electrical energy between two or more circuits through electromagnetic induction, typically to change the voltage level. A step-up transformer converts a low AC voltage to a high AC voltage (statement 1), while a step-down transformer converts a high AC voltage to a low AC voltage (statement 2). Transformers only work with alternating current (AC).
Statement 3 is incorrect; converting direct current (DC) to alternating current (AC) is done by an inverter. Statement 4 is incorrect; regulating voltage fluctuations is the function of voltage regulators or stabilizers, although transformers are part of the power distribution system where voltage levels are managed.

17. Transformers are used in between the electric power stations and homes

Transformers are used in between the electric power stations and homes or factories in order to

[amp_mcq option1=”minimize the power loss in transmission cables” option2=”minimize the voltage drop in transmission cables” option3=”minimize the current drop in the transmission cables” option4=”provide constant voltage at the user end” correct=”option1″]

This question was previously asked in
UPSC CAPF – 2011
Download PDFAttempt Online
Transformers are used in between the electric power stations and homes or factories in order to minimize the power loss in transmission cables.
Electric power is transmitted over long distances through cables. These cables have resistance, and power is lost as heat according to the formula P_loss = I² * R, where I is the current and R is the resistance of the cable. To minimize this power loss for a given amount of power being transmitted (P = V * I), the voltage (V) is stepped up to a very high level by transformers at the power station. This high voltage means the current (I) required to transmit the same power is much lower (I = P/V). A lower current significantly reduces the power loss (I²R) in the transmission lines. At the receiving end, transformers step the voltage back down for safe use in homes and factories.
While stepping up voltage does also reduce voltage drop (V_drop = I*R), the primary purpose of high-voltage transmission is the dramatic reduction in power loss, making long-distance transmission economically feasible. Transformers are essential for changing voltage levels efficiently in AC circuits.

18. At the time of short-circuit the current in the circuit,

At the time of short-circuit the current in the circuit,

[amp_mcq option1=”reduces substantially” option2=”does not change” option3=”increases heavily” option4=”varies continuously” correct=”option3″]

This question was previously asked in
UPSC CAPF – 2011
Download PDFAttempt Online
At the time of short-circuit, the current in the circuit increases heavily.
A short circuit occurs when a low-resistance path is established between points in a circuit that are normally at different electrical potentials. According to Ohm’s Law (V = IR, or I = V/R), if the resistance (R) in a circuit decreases significantly while the voltage (V) remains relatively constant, the current (I) must increase dramatically.
This heavy increase in current during a short circuit can cause overheating, damage to wiring and components, and is a common cause of electrical fires. Protective devices like fuses and circuit breakers are designed to detect this surge in current and interrupt the circuit to prevent damage.

19. The AC mains domestic supply current in India changes direction in

The AC mains domestic supply current in India changes direction in every

[amp_mcq option1=”50 s” option2=”$\frac{1}{50}$ s” option3=”100 s” option4=”$\frac{1}{100}$ s” correct=”option4″]

This question was previously asked in
UPSC NDA-2 – 2024
Download PDFAttempt Online
The standard frequency of AC mains supply in India is 50 Hz. The frequency (f) is the number of complete cycles per second. The time period (T) of one complete cycle is given by T = 1/f. For f = 50 Hz, T = 1/50 s. In one complete cycle of alternating current, the current starts from zero, reaches a maximum in one direction, passes through zero, reaches a maximum in the opposite direction, and returns to zero. The current changes direction twice within one complete cycle (at the points where it crosses the zero line). Thus, the time taken for the current to change direction is half the time period. Time for direction change = T/2 = (1/50 s) / 2 = 1/100 s.
The frequency of AC supply is the number of cycles per second. The time period is the duration of one cycle (T = 1/f). AC current changes direction twice in every complete cycle.
In many other countries, such as the USA, the standard AC mains frequency is 60 Hz. In a 60 Hz system, the current changes direction every 1/120 s.

20. An incandescent electric bulb converts 20% of its power consumption in

An incandescent electric bulb converts 20% of its power consumption into light, and the remaining power is dissipated as heat. The bulb’s filament has a resistance of 200 Ω and 2 A current flows through it. If the bulb remains ON for 10 h and the rate of electricity charge is ₹5/unit, then which among the following is the correct amount for the money spent on producing light ?

[amp_mcq option1=”₹5″ option2=”₹6″ option3=”₹7″ option4=”₹8″ correct=”option4″]

This question was previously asked in
UPSC NDA-2 – 2024
Download PDFAttempt Online
The power consumed by the incandescent bulb is calculated using the given resistance and current. The total energy consumed over 10 hours is then calculated. Since only 20% of the power (and thus energy) is converted into light, the energy spent on producing light is 20% of the total energy consumed. The cost is calculated based on the rate per unit (kWh) for this amount of energy.
Power consumed (P) = I²R = (2 A)² * 200 Ω = 4 * 200 = 800 W = 0.8 kW.
Total energy consumed (E_total) = Power * Time = 0.8 kW * 10 h = 8 kWh.
Total cost of electricity = E_total * Rate = 8 kWh * ₹5/kWh = ₹40.
Energy converted into light (E_light) = 20% of E_total = 0.20 * 8 kWh = 1.6 kWh.
Cost spent on producing light = E_light * Rate = 1.6 kWh * ₹5/kWh = ₹8.
Incandescent bulbs are inefficient in converting electrical energy into visible light; a large portion of the energy is dissipated as heat, which is why the question specifies that only 20% is converted to light. The remaining 80% (1.6 kWh * 4 = 6.4 kWh) is dissipated as heat, costing ₹32 (₹40 – ₹8).
Page 1Page 2Page 3Page 4Page 5