Electricity

Solution

Electric current is the flow of electric charge through a conductor. It is measured as the rate of flow of charge, where the SI unit is the ampere (A). The formula to calculate current is I = Q/t, where I is the current, Q is the charge in coulombs, and t is the time in seconds. For example, if a charge of 5 coulombs flows through a conductor in 2 seconds, the current is 2.5 A. Electric current is measured using an ammeter, which is connected in series in the circuit. The direction of conventional current is opposite to the flow of electrons. In real life, electric current powers devices like bulbs and heaters by allowing electrons to flow through them, providing energy for operation.

Solution

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 remains constant. Mathematically, it is expressed as V = IR, where V is the voltage, I is the current, and R is the resistance. This law is fundamental in designing and analyzing electric circuits because it helps in calculating the current, voltage, or resistance when the other two quantities are known. For example, if a resistor of 10 ohms has a voltage of 20 volts across it, the current through it is 2 amperes. Ohm's Law is crucial for understanding how different components in a circuit interact and is widely used in household and industrial electrical systems.

Solution

The resistance of a conductor depends on several factors: 1) Length - Resistance is directly proportional to the length of the conductor. Longer conductors have higher resistance. 2) Cross-sectional area - Resistance is inversely proportional to the cross-sectional area. Thicker wires have lower resistance. 3) Material - Different materials have different resistivities, affecting resistance. For example, copper has lower resistivity than iron. 4) Temperature - For most conductors, resistance increases with temperature. For example, the resistance of a tungsten filament increases as it heats up. Understanding these factors helps in selecting appropriate materials and dimensions for wires in electrical circuits to minimize energy loss and ensure efficient operation.

Solution

In a series connection, resistors are connected end-to-end, and the same current flows through each resistor. The total resistance is the sum of individual resistances (R_total = R1 + R2 + R3). For example, three resistors of 2Ω, 3Ω, and 5Ω in series give a total resistance of 10Ω. In a parallel connection, resistors are connected across the same voltage, and the current divides among them. The total resistance is given by 1/R_total = 1/R1 + 1/R2 + 1/R3. For example, two resistors of 4Ω and 4Ω in parallel give a total resistance of 2Ω. Series connections are used where a single path for current is needed, while parallel connections are used to provide multiple paths, ensuring that if one path fails, others continue to work.

Solution

The heating effect of electric current occurs when electrical energy is converted into heat energy as current flows through a resistor. This is described by Joule's Law: H = I²Rt, where H is the heat produced, I is the current, R is the resistance, and t is the time. Practical applications include electric heaters, toasters, and incandescent bulbs, where the heating effect is utilized to produce warmth or light. For example, an electric iron uses a heating element with high resistance to generate heat for pressing clothes. Safety devices like fuses also rely on the heating effect to melt and break the circuit during overcurrent, preventing damage to appliances. While useful, the heating effect can also lead to energy loss in transmission lines, which is why high-voltage transmission is preferred to reduce current and thus heat loss.

Solution

Electric power is the rate at which electrical energy is consumed or transferred in a circuit. It is calculated using the formula P = VI, where P is power in watts (W), V is voltage in volts (V), and I is current in amperes (A). Other forms of the formula include P = I²R and P = V²/R, derived from Ohm's Law. For example, a device operating at 220V with a current of 5A consumes 1100W of power. The commercial unit of electrical energy is the kilowatt-hour (kWh), which represents the energy consumed by a 1kW device in one hour. Understanding electric power is essential for designing efficient electrical systems and calculating energy consumption for billing and conservation purposes.

Solution

A fuse is a safety device that protects electrical circuits from overcurrent by melting and breaking the circuit when the current exceeds a safe value. It consists of a thin wire made of a material with a low melting point, such as tin or lead. When excessive current flows, the wire heats up and melts, interrupting the circuit and preventing damage to connected devices. Fuses are rated by their current capacity (e.g., 5A, 10A) and are selected based on the normal operating current of the circuit. For example, a 5A fuse is suitable for a circuit that normally carries 4A. Fuses are crucial in preventing fires and equipment damage caused by short circuits or overloads, making them an essential component in household and industrial electrical systems.

Solution

Conductors are materials that allow electric current to flow easily due to the presence of free electrons, such as metals like copper and aluminum. They have low resistivity and are used in wires and cables to transmit electricity. Insulators, on the other hand, resist the flow of electric current because they have very few free electrons. Examples include rubber, glass, and plastic, which are used to coat wires and prevent accidental contact with live parts. The key difference lies in their electrical resistivity: conductors have resistivity in the range of 10^-8 to 10^-6 Ωm, while insulators have resistivity from 10^12 to 10^17 Ωm. This distinction is vital in designing electrical systems to ensure safe and efficient current flow where needed and insulation where necessary.

Solution

An electric bulb works by passing current through a thin filament, which heats up to a high temperature and emits light due to incandescence. Tungsten is used for the filament because it has a high melting point (3380°C), allowing it to remain solid at the high temperatures required for light emission. Additionally, tungsten has high resistivity, which helps in generating sufficient heat. The filament is enclosed in a glass bulb filled with inert gases like argon or nitrogen to prevent oxidation and prolong its life. When current flows through the filament, electrical energy is converted into heat and light. However, most of the energy is lost as heat, making incandescent bulbs less efficient compared to modern alternatives like LEDs, which convert more energy into light.

Solution

The resistance of a wire is given by R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. If the length is doubled, resistance becomes 2R (since R ∝ L). If the diameter is halved, the area becomes A/4 (since A = πr² and r is halved). Thus, resistance becomes 8R (since R ∝ 1/A). Combining both changes, the new resistance is 2 * 8 = 16 times the original resistance. For example, if the original resistance was 5Ω, the new resistance would be 80Ω. This demonstrates how significantly changes in dimensions can affect resistance, which is important in designing electrical components to meet specific requirements.

Solution

Electric current is the flow of electric charge through a conductor. It is measured as the rate of flow of charge, quantified by the equation I = Q/t, where I is the current, Q is the charge, and t is time. The SI unit of current is ampere (A). An ammeter is used to measure the current in a circuit and is always connected in series to ensure the same current flows through it as through the rest of the circuit.

Solution

In series connections, the same current flows through all resistors, and the total potential difference is the sum of individual potential differences across each resistor. The total resistance is the sum of individual resistances. In parallel connections, the potential difference across each resistor is the same, and the total current is the sum of currents through each resistor. The reciprocal of the total resistance is the sum of the reciprocals of individual resistances.

Solution

The heating effect of electric current is the phenomenon where electrical energy is converted into heat energy when current passes through a resistor. The heat produced (H) can be derived using Joule's law: H = I²Rt, where I is the current, R is the resistance, and t is the time. This shows that heat is directly proportional to the square of the current, the resistance, and the time.

Solution

Tungsten is used because it has a high melting point (3380°C) and can withstand the high temperatures needed to emit light. The bulb is filled with inert gases like nitrogen and argon to prevent oxidation of the filament. The design includes a vacuum or gas filling to reduce heat loss through convection and conduction, and the filament is coiled to increase resistance and thus the heat and light produced.

Solution

The resistance (R) of a wire is given by R = ρl/A, where ρ is resistivity, l is length, and A is cross-sectional area. If the diameter is doubled, the area becomes four times larger (since A = πr²), and thus the resistance becomes one-fourth of its original value.

Solution

The resistance of a conductor depends on its length (l), cross-sectional area (A), and the material's resistivity (ρ). The relationship is given by R = ρl/A. Resistance increases with length and resistivity but decreases with an increase in cross-sectional area.

Solution

A fuse is a safety device that protects electrical circuits from overcurrent. It contains a wire that melts when the current exceeds a certain value, breaking the circuit and preventing damage to appliances or fire hazards. Fuses are rated based on the maximum current they can handle safely.

Solution

Electric power (P) is the rate at which electrical energy is consumed. It can be expressed as P = VI. Using Ohm's law (V = IR), it can also be written as P = I²R or P = V²/R. This shows that power depends on the square of the current or voltage and is directly proportional to resistance in the first form and inversely in the second.

Solution

The 100 W bulb draws more current because power (P) is directly proportional to current (I) for a given voltage (P = VI). Using I = P/V, the 100 W bulb has I = 100/220 ≈ 0.45 A, and the 60 W bulb has I = 60/220 ≈ 0.27 A. Thus, the higher power rating indicates higher current draw.

Solution

The commercial unit of electrical energy is the kilowatt-hour (kWh). To calculate the cost: Energy consumed = Power × Time = 400 W × 8 hours/day × 30 days = 96 kWh. Cost = 96 kWh × Rs 3.00/kWh = Rs 288.00.

Solution

The resistance of a conductor generally increases with temperature due to the increased vibration of atoms, which impedes the flow of electrons. This phenomenon is crucial in devices like incandescent bulbs, where the filament's resistance increases as it heats up, affecting its brightness and efficiency. However, in semiconductors, resistance decreases with temperature, showcasing the material-dependent nature of this relationship.

Solution

Parallel connections allow each appliance to operate independently at the same voltage, ensuring that if one fails, others continue to work. This setup also enables the drawing of higher currents as needed. However, it requires more wiring and can lead to higher energy consumption if not managed properly, as each appliance draws current based on its resistance.

Solution

The equivalent resistance (R_p) for three resistors (R1, R2, R3) in parallel is given by 1/R_p = 1/R1 + 1/R2 + 1/R3. This configuration decreases the total resistance, allowing a higher total current to flow through the circuit compared to a series connection, as the current has multiple paths to take, reducing the overall resistance to flow.

Solution

Thicker wires have lower resistance due to a larger cross-sectional area, reducing power loss (P = I²R) over long distances. Longer wires increase resistance, leading to higher power loss. Thus, for efficient power transmission, thicker and shorter wires are preferred, but practical constraints like cost and space must be considered.

Solution

Fuses protect circuits by melting and breaking the circuit when current exceeds a safe level, preventing damage or fire. Selecting a fuse with a rating slightly higher than the normal operating current ensures it doesn't blow during regular use but protects against surges. Incorrect ratings can lead to either frequent disruptions or failure to protect.

Solution

In series, the same current flows through all resistors, leading to uniform heating proportional to each resistor's resistance. In parallel, the current divides, with more current flowing through lower resistance paths, causing uneven heating. Total power dissipation is higher in parallel due to lower equivalent resistance, leading to more total heat generated.

Solution

Tungsten has a high melting point (3380°C) and high resistivity, making it ideal for filaments that must reach high temperatures to emit light without melting. Its ability to retain strength at high temperatures and low vapor pressure also prolongs the bulb's life, despite the higher cost compared to other materials.

Solution

Energy = Power × Time = 2 kW × 2 h = 4 kWh. In joules, 4 kWh = 4 × 3.6 × 10^6 J = 14.4 × 10^6 J. kWh is practical for billing as it represents large energy quantities, while joules are fundamental units useful in scientific calculations, linking electrical energy to other forms of energy.

Solution

Using a wire with insufficient diameter increases its resistance, leading to excessive heat generation (P = I²R) and potential melting or fire. It also causes significant voltage drops across the wire, reducing the efficiency and performance of connected devices, and may trip circuit breakers or blow fuses due to overcurrent.

Solution

To achieve a desired equivalent resistance, combine resistors in series to add their resistances (R_total = R1 + R2) or in parallel to reduce it (1/R_total = 1/R1 + 1/R2). For example, to get 4Ω from 6Ω resistors, connect two in parallel (1/R_total = 1/6 + 1/6 → R_total = 3Ω) and then add another in series (3Ω + 6Ω = 9Ω isn't desired, so adjust combinations). The exact configuration depends on available resistors and the target resistance, requiring iterative testing or algebraic solving.