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Electricity is one of the most important phenomena in our daily lives. From lighting our homes to powering smartphones and electric vehicles, understanding electricity is essential for grasping how the modern world functions.
At its core, electricity involves the movement of tiny particles called electric charges. These charges create currents that carry energy through wires and devices, enabling them to operate.
This chapter will take you step-by-step through the fundamental concepts of electricity, focusing on building your problem-solving skills needed for competitive exams. You'll learn about electric charges, how currents flow, key laws like Ohm's and Kirchhoff's, and how to analyze electrical circuits-both simple and complex.
By the end, you will confidently apply formulas to solve questions related to power consumption, circuit design, and electrical safety, all using the metric system and examples relevant to our everyday experience.
Electric charge is a fundamental property of certain particles. It can be either positive or negative. For example, protons in an atom carry a positive charge, while electrons carry a negative charge.
Objects become electrically charged when they gain or lose electrons. For instance, rubbing a balloon on your hair transfers electrons, making the balloon negatively charged and your hair positively charged.
Electric current is the flow of electric charge through a material. In metals like copper, it is usually the electrons that move, carrying this charge. The amount of current is measured in amperes (A), indicating how many charges pass a point every second.
In actual physics and electronics, electrons move from the negative terminal to the positive terminal. However, by historical convention, conventional current is considered to flow from the positive terminal to the negative terminal.
This is shown in the diagram below, where the small circles represent electrons moving opposite to the red arrow indicating conventional current direction.
Any electrical circuit involves three important quantities:
Ohm's Law states that, for many materials, the voltage across a resistor is directly proportional to the current flowing through it. This relationship is given by the formula:
\( V = IR \)
Here, \(V\) is voltage in volts, \(I\) is current in amperes, and \(R\) is resistance in ohms.
This means if you increase voltage, current increases proportionally; if resistance increases, current decreases for the same voltage.
Resistance depends on several factors:
This relationship is captured mathematically by resistivity (ρ), a property specific to each material:
\( R = \rho \frac{L}{A} \)
Where:
Electric circuits often have multiple resistors connected together. The way resistors are connected affects the total resistance of the circuit and how current flows.
Resistors are connected one after another in a single path. The current flowing through each resistor is the same, but the voltage divides among the resistors.
The total resistance in series is the sum of all resistors:
\( R_{total} = R_1 + R_2 + \cdots + R_n \)
Resistors are connected across the same two points, creating multiple paths for current. The voltage across each resistor is the same, but current divides among branches.
The total resistance is found using the reciprocal sum formula:
\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \]
Because of multiple paths, the total resistance in parallel is always less than the smallest individual resistor.
When electric current flows through a device, it does work or produces heat or light. This work done or energy conversion is called electrical power.
Power is the rate at which energy is transferred. The basic formula for electrical power is:
\( P = VI \)
Where \(P\) is power in watts (W), \(V\) is voltage in volts, and \(I\) is current in amperes.
Using Ohm's Law (\(V=IR\)), power can also be expressed as:
\( P = I^2 R \quad \text{or} \quad P = \frac{V^2}{R} \)
The total electrical energy consumed (used) over time \(t\) is:
\( E = Pt \)
Energy is measured in joules (J) if time is in seconds, or more practically in kilowatt-hours (kWh) when time is in hours. 1 kWh equals 3,600,000 joules.
Electricity bills in India are calculated based on energy consumed (in kWh) multiplied by the tariff per unit:
\( \text{Cost} = \text{Energy (kWh)} \times \text{Tariff (INR/kWh)} \)
| Formula | Variables | Units | Use |
|---|---|---|---|
| \(P = VI\) | \(P\) = power, \(V\) = voltage, \(I\) = current | W, V, A | Calculate power given voltage and current |
| \(P = I^2 R\) | \(P\), \(I\), \(R\) | W, A, Ω | Calculate power using current and resistance |
| \(P = \frac{V^2}{R}\) | \(P\), \(V\), \(R\) | W, V, Ω | Calculate power using voltage and resistance |
| \(E = Pt\) | \(E\) = energy, \(P\) = power, \(t\) = time | J or kWh, W, s or h | Energy consumed over given time |
| \(\text{Cost} = \text{Energy} \times \text{Tariff}\) | Energy (kWh), tariff (INR/kWh) | INR | Calculate total electricity bill |
Step 1: Write down known values: \(V = 12\,V\), \(R = 4\,\Omega\).
Step 2: Use Ohm's Law formula: \(I = \frac{V}{R}\).
Step 3: Calculate current: \(I = \frac{12}{4} = 3\, A\).
Answer: The current flowing through the resistor is 3 amperes.
Step 1: Write down the resistances: \(R_1 = 2\,\Omega\), \(R_2 = 3\,\Omega\), \(R_3 = 5\,\Omega\).
Step 2: Use series formula: \(R_{total} = R_1 + R_2 + R_3 = 2 + 3 + 5\).
Step 3: Calculate sum: \(R_{total} = 10\, \Omega\).
Answer: Total resistance in series is 10 Ω.
Step 1: Power rating \(P = 1500\,W = 1.5\,kW\), time \(t = 3\,h\), tariff = 8 INR/kWh.
Step 2: Calculate energy used: \(E = P \times t = 1.5 \times 3 = 4.5\, kWh\).
Step 3: Calculate cost: \(\text{Cost} = 4.5\, \text{kWh} \times 8\, \text{INR/kWh} = 36\, \text{INR}\).
Answer: Energy consumed is 4.5 kWh and cost is Rs.36.
Step 1: Find total resistance using parallel formula:
\[ \frac{1}{R_{total}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = 1 + 0.5 + 0.333 = 1.833 \]
Step 2: Calculate \(R_{total}\):
\[ R_{total} = \frac{1}{1.833} \approx 0.545\, \Omega \]
Step 3: Calculate total current using \(I = \frac{V}{R}\):
\[ I = \frac{12}{0.545} \approx 22.02\, A \]
Step 4: Find current through each resistor:
Total current \(= 12 + 6 + 4 = 22\, A\) (matches total current calculated)
Answer: Total current supplied is approximately 22 A; individual branch currents are 12 A, 6 A, and 4 A.
Step 1: Identify loops and assign currents \(I_1\) and \(I_2\).
Step 2: Apply Kirchhoff's Voltage Law (KVL) to the first loop (clockwise direction):
\[ 10 - 2I_1 - 3(I_1 - I_2) = 0 \]
Step 3: Simplify first loop equation:
\[ 10 - 2I_1 - 3I_1 + 3I_2 = 0 \implies 10 = 5I_1 - 3I_2 \]
Step 4: Apply KVL to second loop:
\[ 5 - 5I_2 - 3(I_2 - I_1) = 0 \]
Step 5: Simplify second loop equation:
\[ 5 - 5I_2 - 3I_2 + 3I_1 = 0 \implies 5 = 8I_2 - 3I_1 \]
Step 6: Rearrange both equations:
\[ 5I_1 - 3I_2 = 10 \quad (1) \]
\[ -3I_1 + 8I_2 = 5 \quad (2) \]
Step 7: Solve simultaneous equations:
Multiply (1) by 8 and (2) by 3:\[ 40I_1 - 24I_2 = 80 \]
\[ -9I_1 + 24I_2 = 15 \]
Add both:
\[ 31I_1 = 95 \implies I_1 = \frac{95}{31} \approx 3.06\, A \]
Substitute \(I_1\) into (1):
\[ 5 \times 3.06 - 3I_2 = 10 \implies 15.3 - 3I_2 = 10 \implies 3I_2 = 5.3 \implies I_2 \approx 1.77\, A \]
Answer: The currents are approximately \(I_1 = 3.06\, A\) and \(I_2 = 1.77\, A\).
When to use: Quickly simplifying circuit resistance.
When to use: Avoiding unit mistakes in formula application.
When to use: When one electrical parameter is missing.
When to use: Ensuring correct energy calculations for billing.
When to use: Clarifying circuit analysis problems.
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