DC Circuits & Kirchhoff's Laws: Physics Notes for AP EAMCET

DC Circuits & Kirchhoff’s Laws

(for AP EAMCET – Physics)

1. Simple Explanation

In a direct‑current (DC) circuit all voltages and currents are steady (time‑independent).
Kirchhoff’s Laws are the two fundamental rules that let us write equations for any such circuit:

Law	Statement	Symbolic Form
Kirchhoff’s Current Law (KCL)	The algebraic sum of currents entering a node equals zero.	(\displaystyle \sum I_{\text{in}} = \sum I_{\text{out}})
Kirchhoff’s Voltage Law (KVL)	The algebraic sum of all voltages around any closed loop is zero.	(\displaystyle \sum V_{\text{across elements}} = 0)

Algebraic sum means we treat currents leaving the node as negative and voltages that drop as negative.

2. Key Points

#	Concept	Why It Matters
1	Nodes & Loops	Identify nodes (points where two or more elements meet) and loops (closed paths).
2	Sign Conventions	Positive current flows from higher to lower potential; positive voltage drop is across an element when moving from + to –.
3	Resistors	(V = IR). Always linear.
4	Series vs Parallel	Series: same current, voltages add. Parallel: same voltage, currents add.
5	Voltage Divider	(V_{\text{out}} = V_{\text{in}}\frac{R_{\text{out}}}{R_{\text{total}}}).
6	Current Divider	(I_{\text{branch}} = I_{\text{total}}\frac{R_{\text{other}}}{R_{\text{branch}}+R_{\text{other}}}).
7	Loop & Nodal Analysis	Choose independent loops or nodes to write KVL/KCL equations.
8	Number of Equations	For n unknowns, you need n independent equations. KVL gives one per independent loop, KCL gives one per independent node.
9	Matrix Methods	For complex circuits, write equations in matrix form (AX = B) and solve.
10	Common Pitfalls	- Forgetting sign conventions. - Mixing up series/parallel relationships. - Not checking that equations are independent.

3. Worked Example

Problem

Find the currents (I_1) and (I_2) in the circuit below.

          +---[R1 = 10 Ω]---+---[R2 = 20 Ω]---+
          |                 |                 |
    V1=12V  |                 |      V2=5V    |
          |                 |                 |
          +---[R3 = 30 Ω]---+---[R4 = 15 Ω]---+

All resistors are in series with each other, but the two voltage sources are connected in parallel.

Step‑by‑Step

#	Action	Equation	Notes
1	Label currents	Let (I_1) flow from (V_1) through (R_1) and (R_2). Let (I_2) flow from (V_2) through (R_3) and (R_4).	Currents are independent because the sources are in parallel.
2	Write KVL for loop of (V_1)	(12,\text{V} - I_1(10) - I_1(20) = 0)	(I_1) flows through (R_1) and (R_2).
3	Simplify	(12 = 30I_1) → (I_1 = 0.4,\text{A})
4	Write KVL for loop of (V_2)	(5,\text{V} - I_2(30) - I_2(15) = 0)	(I_2) flows through (R_3) and (R_4).
5	Simplify	(5 = 45I_2) → (I_2 = 0.111,\text{A})
6	Check node	At the central node, currents leave through (R_2) and (R_3) and return through (R_4). Since sources are independent, no extra equation needed.

Results

(I_1 = 0.4 \text{ A}) (through (R_1) and (R_2))
(I_2 = 0.111 \text{ A}) (through (R_3) and (R_4))

If you had a single voltage source powering the series chain (R_1)–(R_2)–(R_3)–(R_4), you would use one KVL equation:
(V = I(R_1+R_2+R_3+R_4)).

4. Quick Revision Summary

Topic	Key Take‑away
KCL	Sum of currents at a node = 0.
KVL	Sum of voltage drops in a loop = 0.
Series	(I_{\text{same}}). (V_{\text{total}} = \sum V).
Parallel	(V_{\text{same}}). (I_{\text{total}} = \sum I).
Voltage Divider	(V_{\text