Electric Current Archives

121. A 100 W electric bulb is used for 10 hours a day. How many units of en

A 100 W electric bulb is used for 10 hours a day. How many units of energy are consumed in 30 days ?

[amp_mcq option1=”1″ option2=”10″ option3=”30″ option4=”300″ correct=”option3″]

This question was previously asked in

UPSC CDS-2 – 2019

The correct answer is 30 units.

Energy consumed is calculated as Power × Time. Power is measured in Watts (W) or Kilowatts (kW), and Time is measured in hours (h). Energy is typically measured in Watt-hours (Wh) or Kilowatt-hours (kWh). One unit of electrical energy is equal to 1 kWh.

Given power = 100 W = 0.1 kW (since 1 kW = 1000 W).
The bulb is used for 10 hours a day for 30 days.
Total time = 10 hours/day × 30 days = 300 hours.
Energy consumed = Power (kW) × Total Time (h)
Energy consumed = 0.1 kW × 300 h = 30 kWh.
Since 1 unit = 1 kWh, the energy consumed is 30 units.

122. A circuit has a fuse having a rating of 5 A. What is the maximum numbe

A circuit has a fuse having a rating of 5 A. What is the maximum number of 100 W-220 V bulbs that can be safely connected in parallel in the circuit?

[amp_mcq option1=”20″ option2=”15″ option3=”11″ option4=”10″ correct=”option4″]

This question was previously asked in

UPSC CDS-2 – 2018

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A maximum of 11 bulbs can be safely connected in parallel in the circuit.

The circuit has a fuse rating of 5 A, which means the total current drawn by all connected devices must not exceed 5 A to prevent the fuse from blowing. Each bulb is rated at 100 W at 220 V. The current drawn by a single bulb is calculated using the formula P = V * I, so I = P/V.
Current per bulb = 100 W / 220 V = 10/22 A = 5/11 A.
When bulbs are connected in parallel, the voltage across each bulb is the same (220 V), and the total current drawn from the source is the sum of the currents drawn by each bulb. Let N be the number of bulbs connected. The total current is N * (5/11) A.
For safe operation, the total current must be less than or equal to the fuse rating: N * (5/11) A ≤ 5 A.
N * 5 ≤ 5 * 11
N ≤ 11.
Since the number of bulbs must be a whole number, the maximum number of bulbs that can be safely connected in parallel is 11.

Fuses are safety devices designed to protect electrical circuits from excessive current that could cause overheating and potentially fire. They are intentionally the weakest point in the circuit and melt, breaking the connection, if the current exceeds their rated value. Parallel connections are common for household appliances like lights, allowing each device to operate independently at the full circuit voltage.

123. Which one of the following does not convert electrical energy into l

Which one of the following does not convert electrical energy into light energy?

[amp_mcq option1=”A candle” option2=”A light-emitting diode” option3=”A laser” option4=”A television set” correct=”option1″]

This question was previously asked in

UPSC CDS-2 – 2018

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The question asks which option does *not* convert electrical energy into light energy.
A) A candle: A candle produces light through the process of combustion, where chemical energy stored in the wax is converted into heat and light energy. It does not use electrical energy.
B) A light-emitting diode (LED): LEDs are semiconductor devices that convert electrical energy directly into light through electroluminescence.
C) A laser: Lasers produce light through stimulated emission. Many types of lasers, particularly semiconductor lasers, are powered by electrical energy. Other lasers may be powered by optical energy, but the principle is often related to electrical excitation or pumping.
D) A television set: A television set’s display screen (whether CRT, LCD, LED, or Plasma) converts electrical energy into light to form images.
Therefore, a candle is the only option that does not convert electrical energy into light energy.

– Identify the energy conversion process in each device.
– Electrical energy to light energy conversion is the key process to look for.
– Understand the working principle of each listed item.

Incandescent bulbs, fluorescent lamps, and gas discharge lamps are other common examples of devices that convert electrical energy into light energy. Photosynthesis in plants converts light energy into chemical energy.

124. After using for some time, big transformers get heated up. This is due

After using for some time, big transformers get heated up. This is due to the fact that

1. current produces heat in the transformers
2. hysteresis loss occurs in the transformers
3. liquid used for cooling gets heated

Select the correct answer using the code given below.

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

This question was previously asked in

UPSC CDS-2 – 2017

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Transformers get heated up due to current producing heat in the windings (copper loss), hysteresis loss in the core, and the liquid used for cooling getting heated as it absorbs these losses.

The main reasons for heating in a transformer are power losses occurring during its operation. These include copper losses (I²R losses) in the windings due to the resistance of the copper wires, and iron losses (core losses) in the core material. Iron losses consist of hysteresis loss (energy dissipated due to the repeated magnetization and demagnetization of the core by the alternating magnetic field) and eddy current loss (induced circulating currents in the core causing resistive heating). Statement 1 correctly identifies copper loss. Statement 2 correctly identifies hysteresis loss, which is one component of iron loss.

Large transformers often use oil or other liquids for cooling. This liquid absorbs the heat generated by the copper and iron losses and circulates, allowing the heat to be dissipated, usually through radiators. The fact that the liquid gets heated (Statement 3) is a direct consequence of the heat generated by losses (1 and 2, plus eddy currents) being transferred to the coolant. While not a primary *source* of heat generation like 1 and 2, the heating of the cooling liquid is an integral part of the thermal behavior and the overall “heated up” state of a large, operating transformer. Thus, all three facts listed contribute to understanding why big transformers get heated up.

125. An electron and a proton starting from rest are accelerated through a

An electron and a proton starting from rest are accelerated through a potential difference of 1000 V. Which one of the following statements in this regard is correct?

[amp_mcq option1=”The kinetic energy of both the particles will be different.” option2=”The speed of the electron will be higher than that of the proton.” option3=”The speed of the proton will be higher than that of the electron.” option4=”The speed of the electron and the proton will be equal.” correct=”option2″]

This question was previously asked in

UPSC CDS-2 – 2017

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The speed of the electron will be higher than that of the proton.

When an electron and a proton, both starting from rest, are accelerated through the same potential difference V, they gain the same kinetic energy (KE). The kinetic energy gained is equal to the work done by the electric field, which is qV, where q is the charge of the particle and V is the potential difference. The magnitude of charge is the same for both electron and proton (|q_e| = |q_p|). So, KE_electron = KE_proton = |q|V.

Kinetic energy is given by KE = 1/2 * mv², where m is the mass and v is the speed. Since KE is the same for both particles, and the mass of the electron (m_e) is significantly less than the mass of the proton (m_p) (m_e ≈ m_p/1836), for the equation 1/2 * m_e * v_e² = 1/2 * m_p * v_p² to hold, the velocity of the lighter particle (electron) must be greater than the velocity of the heavier particle (proton). Thus, v_e > v_p.

126. The r.m.s. potential difference between the red live wire and black ne

The r.m.s. potential difference between the red live wire and black neutral wire in Indian domestic electric supply is

[amp_mcq option1=”160 V” option2=”220 V” option3=”300 V” option4=”410 V” correct=”option2″]

This question was previously asked in

UPSC CDS-1 – 2023

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The standard r.m.s. (root mean square) voltage for domestic electric supply in India is 220 volts. This is the effective voltage that determines the power delivered to electrical appliances. The peak voltage is higher, approximately $220 \times \sqrt{2} \approx 311$ volts.

– The voltage of AC power is usually specified by its r.m.s. value.
– The standard domestic supply voltage in India is 220-240 V AC at 50 Hz.

Different countries use different standard voltages and frequencies for domestic power supply (e.g., 120 V at 60 Hz in North America, 230 V at 50 Hz in much of Europe). The red wire is typically the live or phase wire, and the black wire is the neutral wire. The potential difference exists between the live and neutral wires.

127. Three resistors of resistances 11 $\Omega$, 22 $\Omega$ and 33 $\Omega

Three resistors of resistances 11 $\Omega$, 22 $\Omega$ and 33 $\Omega$ are connected in parallel. Their equivalent resistance is equal to

[amp_mcq option1=”66 $\Omega$” option2=”22 $\Omega$” option3=”12 $\Omega$” option4=”6 $\Omega$” correct=”option4″]

This question was previously asked in

UPSC CDS-1 – 2023

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The equivalent resistance ($R_{eq}$) of resistors connected in parallel is given by the formula:
$1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3 + …$
Given resistances are $R_1 = 11 \, \Omega$, $R_2 = 22 \, \Omega$, and $R_3 = 33 \, \Omega$.
$1/R_{eq} = 1/11 + 1/22 + 1/33$
To add these fractions, find a common denominator, which is 66.
$1/R_{eq} = (6 \times 1)/(6 \times 11) + (3 \times 1)/(3 \times 22) + (2 \times 1)/(2 \times 33)$
$1/R_{eq} = 6/66 + 3/66 + 2/66$
$1/R_{eq} = (6 + 3 + 2) / 66 = 11 / 66$
$R_{eq} = 66 / 11 = 6 \, \Omega$.

– For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.
– The equivalent resistance in a parallel circuit is always less than the smallest individual resistance. In this case, 6 $\Omega$ is less than 11 $\Omega$.

Resistors can be connected in two main ways: series and parallel. In a series connection, resistances add up ($R_{eq} = R_1 + R_2 + R_3 + …$). In a parallel connection, the voltage across each resistor is the same, and the total current is the sum of currents through each resistor.

128. A 100 W electric bulb is used for 10 hours a day. How many units of el

A 100 W electric bulb is used for 10 hours a day. How many units of electrical energy are consumed by the bulb in 3 days? (1 unit = 1 kWh)

[amp_mcq option1=”3.00″ option2=”1.08″ option3=”2.16″ option4=”0.33″ correct=”option1″]

This question was previously asked in

UPSC CDS-1 – 2023

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The energy consumed by an electrical appliance is calculated by multiplying its power rating by the time it is used.
Power (P) = 100 W.
Time used per day = 10 hours.
Total time used in 3 days = 10 hours/day × 3 days = 30 hours.
Energy consumed (in Watt-hours) = P × Time = 100 W × 30 hours = 3000 Wh.
To convert Watt-hours to kilowatt-hours (units), divide by 1000:
Energy consumed (in kWh) = 3000 Wh / 1000 Wh/kWh = 3 kWh.
Since 1 unit = 1 kWh, the energy consumed is 3 units.

Electrical energy consumed is calculated as Power × Time. Energy unit (kWh) = Power (kW) × Time (h).

Electricity meters in homes measure energy consumption in kilowatt-hours (kWh), which are commonly referred to as ‘units’. The cost of electricity is usually charged per unit consumed.

129. Which one of the following statements is not correct ?

Which one of the following statements is not correct ?

[amp_mcq option1=”An electric motor converts electrical energy into mechanical energy.” option2=”An electric generator works on the principle of electromagnetic induction.” option3=”The magnetic field at the centre of a long circular coil carrying current will be parallel straight lines.” option4=”A wire with a green insulation is usually the live wire of an electric supply.” correct=”option4″]

This question was previously asked in

UPSC CDS-1 – 2022

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Statement D is incorrect because a wire with green (or green/yellow) insulation is typically the earth wire in an electric supply, not the live wire.

Standard electrical wiring colour codes are used to identify the different wires in a circuit: Live, Neutral, and Earth. These colours vary by region, but green or green/yellow is almost universally used for the Earth wire, while red or brown is common for the Live wire, and black or blue is common for the Neutral wire.

Statement A is correct as electric motors convert electrical energy into mechanical energy using the force on a current-carrying conductor in a magnetic field. Statement B is correct as electric generators produce electric current based on the principle of electromagnetic induction, where a changing magnetic field induces an electric current. Statement C is correct; inside a long circular coil (solenoid) carrying current, the magnetic field lines are parallel straight lines, indicating a uniform field.

130. 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=”keeps on fluctuating.” correct=”option3″]

This question was previously asked in

UPSC CDS-1 – 2022

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A short-circuit occurs when a low-resistance path is created between two points in an electric circuit that are normally at different potentials, often bypassing the intended load.

According to Ohm’s Law, the current (I) flowing through a circuit is directly proportional to the voltage (V) across it and inversely proportional to the resistance (R): I = V/R. During a short-circuit, the resistance of the path drops significantly, often close to zero. Assuming the voltage source remains relatively constant, a drastic decrease in resistance leads to a massive increase in the current flowing through the circuit.

This heavy increase in current during a short-circuit can cause overheating of wires and components due to the power dissipated (P = I²R), potentially leading to damage, fire, or explosion. Safety devices like fuses and circuit breakers are designed to detect this excessive current and interrupt the circuit, preventing damage.