Electric Current

Which one of the following statements regarding magnetic field is NOT correct ?

[amp_mcq option1=”Magnetic field is a quantity that has direction and magnitude” option2=”Magnetic field lines are closed curves” option3=”Magnetic field lines are open curves” option4=”No two magnetic field lines are found to cross each other” correct=”option3″]

The statement which is NOT correct is (C) Magnetic field lines are open curves. Magnetic field lines form continuous, closed loops. They originate from the North pole and end at the South pole outside the magnet, and continue from the South pole to the North pole inside the magnet.

Magnetic field lines are always closed curves. They provide a visual representation of the direction and strength of a magnetic field.

Statement (A) is correct: Magnetic field is a vector quantity. Statement (B) is correct: Magnetic field lines are closed curves. Statement (D) is correct: Magnetic field lines never cross each other because the magnetic field at any point has a unique direction.

The cost of energy to operate an industrial refrigerator that consumes 5 kW power working 10 hours per day for 30 days will be
(Given that the charge per kW.h of energy = ₹ 4)

[amp_mcq option1=”₹ 600″ option2=”₹ 6,000″ option3=”₹ 1,200″ option4=”₹ 1,500″ correct=”option2″]

The correct answer is (B) ₹ 6,000. First, calculate the total energy consumed. Power is 5 kW. The refrigerator works 10 hours per day for 30 days. Total hours = 10 hours/day * 30 days = 300 hours. Total energy consumed = Power * Time = 5 kW * 300 hours = 1500 kWh. The cost per kWh is ₹ 4. Total cost = 1500 kWh * ₹ 4/kWh = ₹ 6000.

Energy consumption is calculated as the product of power (in kilowatts) and time (in hours), resulting in kilowatt-hours (kWh). The cost is then found by multiplying the total energy consumed by the rate per kWh.

Electrical energy consumption is typically measured in kilowatt-hours (kWh), also known as a ‘unit’ of electricity. This is the standard unit used by electricity meters and utility companies for billing purposes.

Which one of the following statements regarding cathode rays is not correct ?

[amp_mcq option1=”Cathode ray particles are electrons.” option2=”Cathode ray particles start from anode and move towards cathode.” option3=”In the absence of electrical and magnetic fields, cathode rays travel in straight lines.” option4=”Television picture tubes are cathode ray tubes.” correct=”option2″]

The statement that is NOT correct regarding cathode rays is B) Cathode ray particles start from anode and move towards cathode.

Cathode rays are streams of negatively charged particles (electrons). In a discharge tube, these electrons are emitted from the cathode (the negative electrode) and are accelerated towards the anode (the positive electrode) by the electric field between them. Therefore, cathode rays move from the cathode towards the anode, not the other way around.

Statement A is correct: Cathode ray particles are indeed electrons. Statement C is correct: In the absence of external electric or magnetic fields, cathode rays travel in straight lines. Statement D is correct: Older style Television picture tubes (CRTs – Cathode Ray Tubes) and computer monitors utilized controlled beams of cathode rays to create images on a phosphorescent screen.

Water is heated with a coil of resistance R connected to domestic supply. The rise of temperature of water will depend on
1. Supply voltage.
2. Current passing through the coil.
3. Time for which voltage is supplied.
Select the correct answer from among the following :

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

The correct answer is A) 1, 2 and 3.

When a resistance coil is connected to a domestic supply, it generates heat according to Joule’s law of heating. The heat produced (H) is given by $H = I^2 R t$, where I is the current, R is the resistance, and t is the time. Using Ohm’s law ($V = IR$), this can also be expressed as $H = \frac{V^2}{R} t$ or $H = V I t$. The rise in temperature of the water is proportional to the heat supplied to it. Therefore, the rise in temperature depends on the supply voltage (V), the current passing through the coil (I), and the time for which the voltage is supplied (t). All three factors influence the amount of heat generated and transferred to the water.

The specific heat capacity and mass of the water also influence the final temperature rise, but the question asks what the heat *production* (and thus potential temperature rise) depends on from the given options related to the electrical supply and coil. The supply voltage and the resistance of the coil determine the current ($I = V/R$). So, if voltage and time are given, the current is implicitly determined by the resistance, and heat depends on V, R, and t. If voltage and current are given, resistance is implicitly determined ($R=V/I$), and heat depends on V, I, and t. If current, resistance, and time are given, heat is directly $I^2 R t$. Since the options list voltage, current, and time separately, it means that the heat produced (and temperature rise) is a function of all these parameters in one form of the heat equation or another.

Consider the following part of an electric circuit :
[Image of circuit diagram is implied here]
The total electrical resistance in the given part of the electric circuit is

[amp_mcq option1=”$\frac{15}{8}$ ohm” option2=”$\frac{15}{7}$ ohm” option3=”15 ohm” option4=”$\frac{17}{3}$ ohm” correct=”option1″]

While the circuit diagram is not provided in the text, assuming a common configuration that results in one of the given simple fractional values, let’s test a common parallel combination. If the circuit consists of two resistors with resistances 3 ohm and 5 ohm connected in parallel, the total electrical resistance ($R_{total}$) is given by the formula for parallel resistors:
$R_{total} = \frac{R_1 \times R_2}{R_1 + R_2}$
Let $R_1 = 3 \, \Omega$ and $R_2 = 5 \, \Omega$.
$R_{total} = \frac{3 \, \Omega \times 5 \, \Omega}{3 \, \Omega + 5 \, \Omega} = \frac{15 \, \Omega^2}{8 \, \Omega} = \frac{15}{8} \, \Omega$.
This result matches option A. This configuration is a plausible intended diagram for such a question structure.

– Resistors in parallel: The reciprocal of the total resistance is the sum of the reciprocals of individual resistances ($1/R_{total} = 1/R_1 + 1/R_2 + \dots$). For two resistors, this simplifies to $R_{total} = (R_1 \times R_2) / (R_1 + R_2)$.
– Assuming common resistor values (like 3 and 5 ohm) and standard circuit configurations (like parallel) often helps in solving such MCQs when the diagram is missing but options are specific.

Series combination of resistors ($R_{total} = R_1 + R_2 + \dots$) results in a resistance greater than any individual resistance. Parallel combination results in a resistance smaller than the smallest individual resistance. In this case, 15/8 = 1.875 is smaller than both 3 and 5 ohm. Other standard configurations like series-parallel combinations could also result in fractional resistances, but the 3||5 parallel arrangement directly yields 15/8.

“The sum of emfs and potential differences around a closed loop equals zero” is a consequence of

[amp_mcq option1=”Ohm’s law.” option2=”Conservation of charge.” option3=”Conservation of momentum.” option4=”Conservation of energy.” correct=”option4″]

Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of the potential differences (voltages) around any closed loop in a circuit is zero. This law is a direct consequence of the principle of conservation of energy. As a unit charge traverses a closed loop and returns to its starting point, the net work done on it by the electric field (and thus the net change in potential energy) must be zero if no energy is gained or lost within the loop from non-electric sources (which is accounted for by EMFs).

– Kirchhoff’s Voltage Law (KVL) is mathematically expressed as $\sum V = 0$ around any closed loop.
– KVL is based on the conservative nature of the electric field and the conservation of energy.
– Potential difference is defined as the change in potential energy per unit charge. Traversing a closed loop means returning to the initial potential, so the total change in potential (and potential energy) is zero.

– Ohm’s law ($V=IR$) is a specific relationship for a resistive component and is used *within* a loop analysis, but KVL is a fundamental principle for the loop itself.
– Conservation of charge is the basis for Kirchhoff’s Current Law (KCL), which deals with currents at a junction.
– Conservation of momentum is a principle applied in mechanics, not directly related to circuit laws in this manner.

Which one of the following statements regarding Ohm’s law is not correct ?

[amp_mcq option1=”Ohm’s law is an assumption that current through a conductor is always directly proportional to the potential difference applied to it.” option2=”A conducting device obeys Ohm’s law when the resistance of a device is independent of magnitude and polarity of applied potential difference.” option3=”A conducting material obeys Ohm’s law when the resistance of material is independent of the magnitude and direction of applied electric field.” option4=”All homogeneous materials obey Ohm’s law irrespective of whether the field is within range or strong.” correct=”option4″]

The correct option is D) All homogeneous materials obey Ohm’s law irrespective of whether the field is within range or strong. This statement is incorrect.

Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature and other physical conditions remain unchanged (V = IR). A material or device is said to be “ohmic” if its resistance R is constant and independent of the applied voltage/current/electric field over a significant range.
Statements A, B, and C describe aspects of Ohm’s law or materials obeying it. Statement A describes Ohm’s law as the proportionality between V and I (though the “always” can be debated depending on interpretation, compared to D it’s less definitively wrong). Statements B and C correctly define an ohmic device/material as one whose resistance is independent of the magnitude and direction of the applied potential difference/electric field. Statement D is a universal claim that is false. Many homogeneous materials (e.g., semiconductors, diodes, electrolytes) do *not* obey Ohm’s law. Even ohmic materials like metals can deviate from Ohm’s law at very high electric fields or extreme temperatures.

Materials that do not obey Ohm’s law are called non-ohmic materials. Their resistance can vary with voltage, current, field strength, temperature, etc. Examples include diodes, transistors, and gas discharge lamps.

A circular coil of radius R having N number of turns carries a steady current I. The magnetic induction at the centre of the coil is 0·1 tesla. If the number of turns is doubled and the radius is halved, which one of the following will be the correct value for the magnetic induction at the centre of the coil?

[amp_mcq option1=”0·05 tesla” option2=”0·2 tesla” option3=”0·4 tesla” option4=”0·8 tesla” correct=”option3″]

The correct value for the magnetic induction at the centre of the coil will be 0.4 tesla.

The magnetic field (B) at the centre of a circular coil is given by the formula $B = (\mu_0 N I) / (2R)$, where $\mu_0$ is the permeability of free space, N is the number of turns, I is the current, and R is the radius of the coil.

Initially, $B_1 = (\mu_0 N_1 I_1) / (2R_1) = 0.1$ T. Let $N_1=N$, $R_1=R$, $I_1=I$.
So, $0.1 = (\mu_0 N I) / (2R)$.
In the new scenario, the number of turns is doubled ($N_2 = 2N$) and the radius is halved ($R_2 = R/2$), while the current remains steady ($I_2 = I$).
The new magnetic field $B_2$ is given by:
$B_2 = (\mu_0 N_2 I_2) / (2R_2) = (\mu_0 (2N) I) / (2(R/2)) = (\mu_0 2N I) / R$.
We can express $B_2$ in terms of $B_1$:
$B_2 = (\mu_0 2N I) / R = 4 \times (\mu_0 N I) / (2R) = 4 \times B_1$.
Substituting the value of $B_1$:
$B_2 = 4 \times 0.1$ T = 0.4 T.

The graphs between current (I) and voltage (V) for three linear resistors 1, 2 and 3 are given below :

If R₁, R₂, and R₃ are the resistances of these resistors, then which one of the following is correct?

[amp_mcq option1=”R₁ > R₂ > R₃” option2=”R₃ < R₁ < R₂" option3="R₃ < R₂ < R₁" option4="R₃ > R₂ > R₁” correct=”option4″]

The correct relationship between the resistances is R₃ > R₂ > R₁.

For a linear resistor obeying Ohm’s Law (V=IR), the relationship between current (I) and voltage (V) is linear. On a graph with Current (I) on the Y-axis and Voltage (V) on the X-axis, the slope of the line represents I/V. According to Ohm’s Law, V/I = R, so I/V = 1/R. Therefore, the slope of the I-V graph is inversely proportional to the resistance (Slope = 1/R).

A steeper slope on this I-V graph indicates a higher value of I for a given V, meaning a higher I/V ratio. Since Slope = 1/R, a higher slope corresponds to a lower resistance.
From the graph, line 1 has the steepest slope, followed by line 2, and line 3 has the shallowest slope.
Slope₁ > Slope₂ > Slope₃
Since Slope = 1/R, this implies:
1/R₁ > 1/R₂ > 1/R₃
Taking the reciprocal reverses the inequality signs:
R₁ < R₂ < R₃ Therefore, the order of resistances from largest to smallest is R₃ > R₂ > R₁.

The connecting cable of electrical appliances like electric iron, water heater or room heater contains three insulated copper wires of three different colours-red, green and black. Which one of the following is the correct colour code?

[amp_mcq option1=”Red-live wire, Green-neutral wire, Black-ground wire” option2=”Red-live wire, Green-ground wire, Black-neutral wire” option3=”Red-live wire, Green-neutral wire, Black-live wire” option4=”Red-ground wire, Green-live wire, Black-neutral wire” correct=”option2″]

The correct colour code among the given options is Red-live wire, Green-ground wire, Black-neutral wire.

Standard colour codes for electrical wiring are used for safety and proper connection. While modern standards in many places use Brown for live, Blue for neutral, and Green/Yellow for earth/ground, older or alternative standards often used Red for live, Black for neutral, and Green for earth/ground. The question uses Red, Green, and Black, indicating reference to such a standard.

In older wiring standards (and sometimes still used in specific applications or regions), the colour coding is typically:
– Live wire: Red (transmits voltage)
– Neutral wire: Black (completes the circuit)
– Earth/Ground wire: Green or Green and Yellow (safety connection to ground)
Based on this established convention using the colours provided, the live wire is Red, the neutral wire is Black, and the ground wire is Green. Option B correctly matches these.