Electricity MCQ Class 10 Quiz

Correct Answer: B. From the positive terminal of the cell to the negative terminal. By convention, the direction of electric current (conventional current) is defined as the direction positive charges would flow, which is from the positive terminal to the negative terminal in an external circuit. This is opposite to the actual flow of electrons.

Correct Answer: B. Coulomb. The coulomb (C) is the standard unit of electric charge in the International System of Units (SI).

Correct Answer: C. Ampere. The ampere (A) is the SI base unit used to measure electric current.

Correct Answer: A. 6 × 10¹⁸. The elementary charge (charge of one electron) is approximately 1.602 × 10⁻¹⁹ C. Therefore, one coulomb contains about 1 ⁄ (1.602 × 10⁻¹⁹) ≈ 6.24 × 10¹⁸ electrons, which is commonly approximated as 6 × 10¹⁸.

Correct Answer: B. Ammeter. An ammeter is designed to measure the rate of flow of electric charge, which is the electric current.

Correct Answer: B. In series. To measure the total current flowing through a part of a circuit, the ammeter must be placed in series, so all the charge flows through it.

Correct Answer: A. A continuous and closed path of an electric current. For a steady current to flow, there must be an unbroken, conducting loop connecting the terminals of the energy source.

Correct Answer: D. Voltmeter. A basic functional circuit needs an energy source (like a battery), a conductive path (wires), and a load (like a resistor or bulb) to consume energy. A switch is for control. A voltmeter measures potential difference and is not required for the circuit to operate.

Correct Answer: B. Electrons. In solid metallic conductors, the charge carriers are the free valence electrons, which can move throughout the metal lattice.

Correct Answer: D. 10 C. Electric charge (Q) is calculated by multiplying the current (I) by the time (t) for which it flows: Q = I × t = 2 A × 5 s = 10 C.

Correct Answer: C. 5 A. Electric current (I) is the rate of flow of charge (Q) over time (t): I = Q ⁄ t = 50 C ⁄ 10 s = 5 A.

Correct Answer: B. 0.2 A. The prefix ‘milli’ (m) represents 10⁻³. So, 200 mA = 200 × 10⁻³ A = 0.2 A.

Correct Answer: A. 5 × 10⁻⁶ A. The prefix ‘micro’ (μ) represents 10⁻⁶. So, 5 μA = 5 × 10⁻⁶ A.

Correct Answer: C. To make or break the flow of current in the circuit. A switch is used to conveniently open (break) or close (make) the conducting path of the circuit.

Correct Answer: C. The current stops flowing, and the bulb does not glow. Turning the switch off creates a break (open circuit) in the path, interrupting the flow of current, which stops the bulb from glowing.

Correct Answer: C. Potential difference. A difference in electric potential (voltage) between two points in a conductor creates an electric field that exerts a force on charge carriers, causing them to move (flow).

Correct Answer: C. Volt. The volt (V) is the SI derived unit for electric potential difference (or voltage).

Correct Answer: C. Battery. A battery (or cell, or power supply) converts chemical energy into electrical energy to maintain a potential difference between its terminals, driving current through an external circuit.

Correct Answer: A. 1 joule of work is done to move a charge of 1 coulomb from one point to another. This is the definition of the volt: 1 V = 1 Joule⁄Coulomb.

Correct Answer: B. 10 joules. Potential difference (V) is defined as energy (E) per unit charge (Q), so E = V × Q. For V=10 V and Q=1 C, the energy is E = 10 V × 1 C = 10 J.

Correct Answer: D. 50 J. Work done (W) in moving a charge (Q) across a potential difference (V) is given by W = Q × V = 5 C × 10 V = 50 J.

Correct Answer: A. 5 V. Potential difference (V) is work done (W) per unit charge (Q): V = W ⁄ Q = 100 J ⁄ 20 C = 5 V.

Correct Answer: A. Directly proportional. This is a statement of Ohm’s Law (V = IR), which applies to many conductors (ohmic conductors) under constant temperature. Voltage is directly proportional to current.

Correct Answer: B. Voltmeter. A voltmeter is designed to measure the electric potential difference (voltage) between two points in a circuit.

Correct Answer: B. In parallel. A voltmeter is connected in parallel (across) the component whose potential difference is being measured.

Correct Answer: A. 9 V. The potential difference (V) provided by the battery is the work done (W) per unit charge (Q): V = W ⁄ Q = 36 J ⁄ 4 C = 9 V.

Correct Answer: D. 24 J. Work done (W) is the product of charge (Q) and potential difference (V): W = Q × V = 2 C × 12 V = 24 J.

Correct Answer: C. To provide a potential difference. The battery acts as the energy source, establishing the potential difference (voltage) that drives the current through the circuit.

Correct Answer: A. It increases. Assuming the conductor obeys Ohm’s Law (V=IR) and its resistance (R) is constant, the potential difference (V) is directly proportional to the current (I). Therefore, if the current increases, the potential difference across the conductor also increases.

Correct Answer: B. Because the battery creates a potential difference across the bulb. This potential difference causes current to flow through the filament. The filament’s resistance converts electrical energy into heat and light (heating effect).

Correct Answer: C. A simplified drawing of an electric circuit using standard symbols for components. Circuit diagrams provide a schematic representation using universally recognized symbols for clarity and ease of understanding.

Correct Answer: B. To represent electrical circuits in a simpler and more understandable way. Standard symbols abstract away physical details, focusing on the connections and function of the circuit.

Correct Answer: B. A closed switch. Based on the symbols provided in this set of questions (specifically comparing with Q39), this symbol represents a closed switch, indicating a complete path.

Correct Answer: A. A resistor. Although a zigzag line (‘—/\/\/\—’) is the standard symbol, in the context of these questions, ‘—•—’ is indicated as representing a resistor.

Correct Answer: A. An ammeter. The letter ‘A’, often enclosed in a circle ‘—(A)—’, is the standard symbol for an ammeter.

Correct Answer: A. A voltmeter. The letter ‘V’, often enclosed in a circle ‘—(V)—’, is the standard symbol for a voltmeter.

Correct Answer: B. ‘—||—’. This symbol strictly represents a single electric cell. A battery (multiple cells) is often shown as ‘—|i|i|—’, but this single cell symbol is frequently used to represent a general DC voltage source or battery in simple diagrams.

Correct Answer: A. Positive and negative terminals of the cell, respectively. In the standard symbol for a cell (‘—||—’), the longer line denotes the positive (+) terminal, and the shorter line denotes the negative (-) terminal.

Correct Answer: B. ‘—/ \’. This symbol shows a break in the line, representing an open switch that interrupts the circuit path.

Correct Answer: B. To make the diagrams universally understandable. Using standardized symbols ensures that engineers and technicians worldwide can interpret circuit diagrams consistently.

Correct Answer: A. Directly proportional. According to Ohm’s Law, for an ohmic conductor at constant temperature, the current is directly proportional to the applied potential difference (I ∝ V).

Correct Answer: A. The opposition offered to the flow of current by the conductor. Resistance quantifies how much a material impedes the flow of electric charge.

Correct Answer: C. Ohm. The ohm (Ω) is the SI unit of electrical resistance, defined as volts per ampere.

Correct Answer: C. The current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature remains constant. Mathematically stated as V=IR or I=V⁄R.

Correct Answer: B. It doubles. From Ohm’s Law (I = V⁄R), if V is doubled and R is constant, I must also double.

Correct Answer: A. It halves. From Ohm’s Law (I = V⁄R), if R is doubled and V is constant, I must become half of its original value.

Correct Answer: B. 2 A. Using Ohm’s Law, I = V⁄R = 10 V ⁄ 5 Ω = 2 A.

Correct Answer: B. 10 Ω. Using Ohm’s Law, R = V⁄I = 20 V ⁄ 2 A = 10 Ω.

Correct Answer: D. A device used to provide a variable resistance. A rheostat is a type of variable resistor, typically used to control the current in a circuit.

Correct Answer: A. To change the current in the circuit. By adjusting the rheostat’s resistance, the total resistance of the circuit changes, which in turn changes the current according to Ohm’s Law (assuming constant voltage).

Correct Answer: B. It decreases. According to Ohm’s Law (I=V⁄R), if the total resistance (R) increases while the voltage (V) remains constant, the current (I) must decrease.

Correct Answer: B. It decreases. The brightness is related to the power dissipated (P=I²R). If the current (I) decreases, the power dissipated as heat and light decreases, making the bulb dimmer.

Correct Answer: B. Because the total potential difference across the circuit increases. Connecting cells in series adds their voltages. With increased total voltage (V) and assuming resistance (R) is constant, the current (I=V⁄R) increases.

Correct Answer: C. 5 Ω. For resistors in series, the equivalent resistance (Rₑ<0xE1><0xB5><0xA1>) is the sum of individual resistances: Rₑ<0xE1><0xB5><0xA1> = R₁ + R₂ = 2 Ω + 3 Ω = 5 Ω.

Correct Answer: C. 2 Ω. For resistors in parallel, 1⁄Rₑ<0xE1><0xB5><0xA1> = 1⁄R₁ + 1⁄R₂. So, 1⁄Rₑ<0xE1><0xB5><0xA1> = 1⁄(4 Ω) + 1⁄(4 Ω) = 2⁄(4 Ω) = 1⁄(2 Ω). Therefore, Rₑ<0xE1><0xB5><0xA1> = 2 Ω.

Correct Answer: B. Length, cross-sectional area, and material. The resistance (R) is given by R = ρL⁄A, where L is length, A is cross-sectional area, and ρ is the resistivity (a property of the material). Temperature also affects resistivity (ρ) and thus resistance.

Correct Answer: A. Directly proportional. Resistance is directly proportional to length (R ∝ L), meaning a longer wire has more resistance, all else being equal.

Correct Answer: B. Inversely proportional. Resistance is inversely proportional to the cross-sectional area (R ∝ 1⁄A), meaning a thicker wire (larger area) has less resistance, all else being equal.

Correct Answer: B. It doubles. Since R ∝ L, doubling L doubles R.

Correct Answer: A. It halves. Since R ∝ 1⁄A, doubling A makes R half its original value.

Correct Answer: A. The resistance of a conductor of unit length and unit cross-sectional area. Resistivity (ρ) is an intrinsic material property defined by R = ρL⁄A. It represents the resistance of a standard cube (e.g., 1m x 1m x 1m) of the material.

Correct Answer: B. Ohm-meter. From ρ = RA⁄L, the units are Ω⋅m² ⁄ m = Ω⋅m.

Correct Answer: D. Nichrome. Silver, copper, and aluminum are excellent conductors with very low resistivity. Nichrome (an alloy of nickel and chromium) has significantly higher resistivity and is used in heating elements.

Correct Answer: C. Because they have low resistivity. Low resistivity minimizes energy loss (P=I²R) due to heat during the transmission of electrical power over long distances.

Correct Answer: A. A good conductor has low resistivity, while a poor conductor has high resistivity. Low resistivity facilitates easy current flow (conductor), while high resistivity impedes it (poor conductor or insulator).

Correct Answer: A. The sum of their individual resistances. Rₑ<0xE1><0xB5><0xA1>(series) = R₁ + R₂ + … + R<0xE2><0x82><0x99>.

Correct Answer: B. The product of their individual resistances divided by their sum. The general formula is 1⁄Rₑ<0xE1><0xB5><0xA1>(parallel) = 1⁄R₁ + 1⁄R₂ + …. For two resistors, this simplifies to Rₑ<0xE1><0xB5><0xA1> = (R₁R₂) ⁄ (R₁ + R₂).

Correct Answer: D. 12 Ω. Rₑ<0xE1><0xB5><0xA1>(series) = 2 Ω + 4 Ω + 6 Ω = 12 Ω.

Correct Answer: C. 1 Ω. 1⁄Rₑ<0xE1><0xB5><0xA1>(parallel) = 1⁄(3 Ω) + 1⁄(3 Ω) + 1⁄(3 Ω) = 3⁄(3 Ω) = 1⁄(1 Ω). Thus, Rₑ<0xE1><0xB5><0xA1> = 1 Ω.

Correct Answer: A. Current. There is only one path for charge flow in a simple series circuit, so the current is the same through all components.

Correct Answer: B. Potential difference. Components connected in parallel share the same two connection points, so the voltage across each component is identical.

Correct Answer: C. Because if one appliance fails, all other appliances in the circuit will also stop working. A break anywhere in a series circuit interrupts the entire path, causing all devices to cease functioning. Also, voltage divides across series components, which isn’t suitable for appliances designed for a standard voltage.

Correct Answer: A. It increases. The total resistance in series is the sum of individual resistances, so adding more resistors always increases the total.

Correct Answer: B. It decreases. Adding parallel paths provides more ways for current to flow, reducing the overall opposition. The equivalent resistance is always less than the smallest individual resistance in the parallel group.

Correct Answer: A. 2.4 Ω. Rₑ<0xE1><0xB5><0xA1> = (R₁R₂) ⁄ (R₁ + R₂) = (4 Ω × 6 Ω) ⁄ (4 Ω + 6 Ω) = 24 Ω² ⁄ 10 Ω = 2.4 Ω.

Correct Answer: B. 1.33 Ω. Rₑ<0xE1><0xB5><0xA1> = (R₁R₂) ⁄ (R₁ + R₂) = (2 Ω × 4 Ω) ⁄ (2 Ω + 4 Ω) = 8 Ω² ⁄ 6 Ω = 4⁄3 Ω ≈ 1.33 Ω.

Correct Answer: C. 6.67 Ω. Rₑ<0xE1><0xB5><0xA1> = (10 Ω × 20 Ω) ⁄ (10 Ω + 20 Ω) = 200 Ω² ⁄ 30 Ω = 20⁄3 Ω ≈ 6.67 Ω.

Correct Answer: C. 1 Ω. 1⁄Rₑ<0xE1><0xB5><0xA1> = 1⁄(2 Ω) + 1⁄(3 Ω) + 1⁄(6 Ω) = (3+2+1)⁄(6 Ω) = 6⁄(6 Ω) = 1⁄(1 Ω). So, Rₑ<0xE1><0xB5><0xA1> = 1 Ω.

Correct Answer: C. Because if one appliance fails, the others continue to work. This provides reliability. Additionally, parallel wiring ensures each appliance receives the full supply voltage (as stated in B), which is necessary for proper operation.

Correct Answer: D. If one appliance fails, the others continue to work in the parallel circuit. This independent operation is a key benefit over series connections for household wiring.

Correct Answer: D. All of the above. Current flow invariably produces heat (P=I²R) and a magnetic field around the conductor. Light is produced if the heating is intense enough (incandescence) or via other mechanisms (e.g., LEDs).

Correct Answer: A. The conversion of electrical energy into heat energy. As charges move through a resistive material, they collide with atoms, transferring energy and increasing the material’s internal energy, which manifests as heat. This is also called Joule heating.

Correct Answer: A. The heat produced in a resistor is directly proportional to the square of the current, the resistance, and the time for which the current flows. The law states that the heat (H) generated is H = I²Rt.

Correct Answer: A. H = I²Rt. This is the mathematical expression of Joule’s law of heating, where H is heat energy, I is current, R is resistance, and t is time.

Correct Answer: B. It quadruples. Since H ∝ I², if I becomes 2I, then H becomes proportional to (2I)² = 4I². The heat produced increases by a factor of 4.

Correct Answer: A. It doubles. Since H ∝ R (assuming I and t are constant), if R becomes 2R, then H becomes 2H. The heat produced doubles.

Correct Answer: C. Electric motor. Electric motors operate based on the magnetic effect of electric current (force on a current-carrying wire in a magnetic field), not the heating effect. Heaters, incandescent bulbs, and fuses directly utilize I²R heating.

Correct Answer: A. It breaks the circuit when excessive current flows through it, due to the melting of a wire with a low melting point. The fuse wire heats up (H=I²Rt) and melts when the current exceeds its rating, creating an open circuit and protecting downstream components.

Correct Answer: B. Because it has a high melting point. Tungsten’s extremely high melting point (approx. 3422 °C) allows the filament to reach the high temperatures needed for incandescence (glowing white-hot) without melting.

Correct Answer: B. Because alloys have a higher melting point than pure metals. Alloys like nichrome are chosen for heating elements because they possess both high resistivity (to generate heat efficiently, P=I²R) and a high melting point to withstand the operating temperatures.

Correct Answer: A. The rate at which electric energy is consumed or dissipated. Power (P) is defined as energy (E) transferred or converted per unit time (t), i.e., P = E⁄t.

Correct Answer: B. Watt. The watt (W) is the SI unit of power, equivalent to one joule per second (1 W = 1 J⁄s).

Correct Answer: A. P = VI. Electric power (P) is the product of the potential difference (V) and the current (I).

Correct Answer: A. P = I²R. Derived from P=VI and Ohm’s law (V=IR).

Correct Answer: A. P = V²⁄R. Derived from P=VI and Ohm’s law (I=V⁄R).

Correct Answer: C. 4840 W. Using P = V²⁄R = (220 V)² ⁄ (10 Ω) = 48400 ⁄ 10 W = 4840 W.

Correct Answer: B. 0.45 A. Using P = VI, the current I = P⁄V = 100 W ⁄ 220 V ≈ 0.4545 A.

Correct Answer: B. A unit of energy. It represents the energy consumed when 1 kilowatt of power is used for 1 hour (Energy = Power × Time).

Correct Answer: A. 3.6 × 10⁶ J. 1 kWh = (1000 W) × (3600 s) = 3,600,000 W⋅s = 3,600,000 J = 3.6 × 10⁶ J.

Correct Answer: C. Kilowatt-hour. Utility companies measure and bill for electrical energy consumption in kilowatt-hours (kWh), often called “units”.