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In electrical engineering, understanding how electrical components connect and interact is fundamental. This section focuses on Direct Current (DC) circuits, particularly on how resistors connect in series and parallel arrangements.
A DC circuit means the current flows in one constant direction, unlike alternating current (AC) which changes direction periodically. DC circuits are common in batteries, mobile devices, and many electronic systems.
We start by defining some basic electrical quantities:
Understanding how voltage, current, and resistance behave in different circuit configurations enables us to analyze and design electrical systems efficiently. Most importantly, mastering these basics will prepare you for more advanced studies and practical engineering challenges.
The foundation of DC circuit analysis lies in fundamental laws that explain how voltage, current, and resistance interact, and how currents and voltages distribute around a circuit.
Ohm's Law states that the voltage \( V \) across a resistor is directly proportional to the current \( I \) flowing through it, with resistance \( R \) being the constant of proportionality:
This simple but powerful law lets us calculate one of the three quantities if the other two are known.
KCL is based on the conservation of electric charge. It states:
"The algebraic sum of currents entering a node (junction) in an electrical circuit is zero."
This means the total current flowing into a point equals the total current flowing out. For example, if 5 A enters a junction, the sum of currents leaving must also be 5 A.
KVL follows from energy conservation and states:
"The algebraic sum of all voltages around any closed loop in a circuit equals zero."
This means if you travel around a loop starting and ending at the same point, the total gains and drops in voltage cancel out.
In a series circuit, components are connected end-to-end, forming a single path for current flow. The current passes through each resistor one after the other.
Key points about series circuits:
The equivalent resistance \( R_{eq} \) for resistors in series is:
Voltage across each resistor can be found using Ohm's law:
\[V_i = I \times R_i\]where \( V_i \) is the voltage drop across the \( i^{th} \) resistor.
In a parallel circuit, components share two common points, creating branches where current can split. Each resistor connects directly across the voltage source.
Key points about parallel circuits:
The equivalent resistance of resistors in parallel is:
The current through each branch can be calculated by:
\[I_i = \frac{V}{R_i}\]where \( I_i \) is the current through resistor \( R_i \) and \( V \) is the common voltage across all branches.
Real-world circuits are often a combination of series and parallel resistors. To analyze these mixed circuits, we break them down step-by-step:
This approach helps solve complex circuits efficiently, especially in competitive exams.
Step 1: Calculate total resistance:
\( R_{eq} = 4 + 6 + 10 = 20 \, \Omega \)
Step 2: Calculate current using Ohm's Law:
\( I = \frac{V}{R_{eq}} = \frac{12}{20} = 0.6\, A \)
Step 3: Calculate voltage drop across each resistor:
Answer: Total resistance = 20 Ω, current = 0.6 A; voltages across resistors are 2.4 V, 3.6 V, and 6.0 V respectively.
Step 1: Voltage across each resistor is same as battery voltage, \( V = 12\, V \).
Step 2: Calculate current through each resistor using Ohm's law:
Step 3: Total current is sum of branch currents:
\( I_{total} = I_1 + I_2 = 4 + 2 = 6\, A \)
Answer: Currents through resistors are 4 A and 2 A; total current supplied by battery is 6 A.
Step 1: Calculate equivalent resistance of parallel part:
\( \frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} = \frac{1}{4} \Rightarrow R_p = 4\, \Omega \)
Step 2: Total resistance of circuit:
\( R_{eq} = R_1 + R_p = 4 + 4 = 8\, \Omega \)
Step 3: Total current from battery:
\( I_{total} = \frac{V}{R_{eq}} = \frac{24}{8} = 3\, A \)
Step 4: Voltage drop across R1:
\( V_1 = I_{total} \times R_1 = 3 \times 4 = 12\, V \)
Step 5: Voltage across parallel branches:
\( V_p = V - V_1 = 24 - 12 = 12\, V \)
Step 6: Calculate current through R2 and R3:
Answer: Equivalent resistance = 8 Ω, total current = 3 A, branch currents: 3 A through R1, 2 A through R2, 1 A through R3.
Step 1: Current through all resistors is 0.6 A (from Example 1).
Step 2: Calculate individual power:
Step 3: Total power dissipated:
\( P_{total} = 1.44 + 2.16 + 3.6 = 7.2\, W \)
Step 4: Input power from battery:
\( P = V \times I = 12 \times 0.6 = 7.2\, W \)
Answer: Power dissipated matches input power (7.2 W), confirming energy balance.
Step 1: Convert power to kilowatts:
\( P = \frac{150}{1000} = 0.15\, kW \)
Step 2: Energy consumed over 5 hours:
\( E = P \times t = 0.15 \times 5 = 0.75\, kWh \)
Step 3: Calculate cost:
\( \text{Cost} = E \times \text{Tariff} = 0.75 \times 8 = Rs.6 \)
Answer: The cost of running the circuit for 5 hours is Rs.6.
When to use: At the start of any circuit problem to avoid confusion and errors.
When to use: For complex circuits where series-parallel reduction alone is not enough.
When to use: During power calculations to save time.
When to use: When only branch current is required and time is limited.
When to use: In cost estimation and billing related questions.
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