DC Circuits & Kirchhoff's Laws

DC Circuits & Kirchhoff’s Laws

(AP EAMCET – Physics – Beginner Notes)

1. What is a DC Circuit?

• DC = Direct Current – a steady flow of electrons.
• Think of a water pipe that always flows in one direction.
• A circuit is a closed loop that lets the current travel from the battery, through wires, and back.

2. Basic Electrical Terms

3. Ohm’s Law – The Simple Rule

V = I × R
(Voltage = Current × Resistance)

• If you know two of V, I, R, you can find the third.
• Example
  • Battery: 9 V
  • Resistor: 3 Ω
  • Find current:
    I = V / R = 9 V / 3 Ω = 3 A

4. Series Circuits

• Components connected end‑to‑end.
• The same current flows through each part.
• Voltage adds up.
• Resistance adds up.

Example

Battery (12 V) → Resistor R1 (2 Ω) → Resistor R2 (4 Ω) → Battery

• Total R = 2 Ω + 4 Ω = 6 Ω
• Current I = 12 V / 6 Ω = 2 A
• Voltage across R1 = I × R1 = 2 A × 2 Ω = 4 V
• Voltage across R2 = 8 V

5. Parallel Circuits

• Components connected side‑by‑side.
• The same voltage appears across each branch.
• Current splits – each branch gets part of the total current.
• Total resistance is less than the smallest resistor.

Example

Battery (12 V) → Junction → R1 (4 Ω) || R2 (6 Ω) → Return to Battery

• Voltage across each resistor = 12 V
• Current through R1 = 12 V / 4 Ω = 3 A
• Current through R2 = 12 V / 6 Ω = 2 A
• Total current = 3 A + 2 A = 5 A

6. Kirchhoff’s Laws – Rules for Complex Circuits

6.1 Kirchhoff’s Current Law (KCL) – “What Goes In, Comes Out”

• At any junction, the sum of currents entering equals the sum of currents leaving.
• Water‑pipe analogy: If 5 L/min of water enters a junction and 3 L/min leaves one branch, 2 L/min must leave the other branch.

Example

      I1
   ────────►
   |        |
   |        |
   |        |
   └───────► I3
      I2

• Let I1 = 4 A, I2 = 3 A.
• KCL: I1 + I2 = I3 → 4 A + 3 A = 7 A (so I3 = 7 A).

6.2 Kirchhoff’s Voltage Law (KVL) – “No Energy is Lost”

• Around any closed loop, the sum of voltage rises and drops is zero.
• Think of a roller coaster: the energy you gain climbing up is exactly the energy you lose going down.

Example

Battery (12 V) → Resistor R1 (2 Ω) → Resistor R2 (4 Ω) → Back to Battery

• Current I = 12 V / (2 Ω + 4 Ω) = 2 A
• Voltage across R1 = I × R1 = 4 V
• Voltage across R2 = I × R2 = 8 V
• KVL check: 12 V (rise) – 4 V – 8 V = 0 V (works!)

7. Quick Problem – Putting It All Together

Problem
A 10 V battery is connected to three resistors:

• R1 = 2 Ω (series)
• R2 = 3 Ω (parallel with R3)
• R3 = 6 Ω

Find the current through each resistor.

Solution

1. Find equivalent resistance of R2‖R3
   [ R_{23} = \frac{1}{\frac{1}{3}+ \frac{1}{6}} = \frac{1}{\frac{1}{2}} = 2,\Omega ]

2. Total resistance (R1 + R23)
   [ R_{\text{total}} = 2,\Omega + 2,\Omega = 4,\Omega ]

3. Total current from battery
   [ I_{\text{total}} = \frac{V_{\text{battery}}}{R_{\text{total}}} = \frac{10,V}{4,\Omega} = 2.5,A ]

4. Current through R1 – same as total current
   [ I_{R1} = 2.5,A ]

5. Voltage across R1
   [ V_{R1} = I_{R1} \times R1 = 2.5,A \times 2,\Omega = 5,V ]

6. Voltage across R2 & R3 (same because they’re parallel)
   [