ElectricalDC CircuitseasyFE ExamCoursework

Node Voltage with Parallel Branches

A $10\ \text{V}$ source connects through a $2\ \Omega$ series resistor to node $A$ .

From node $A$ , two $4\ \Omega$ resistors run in parallel to ground.

Find the voltage at node $A$ .

Given

Hint 1

Simplify before analysing — what do the two parallel resistors reduce to?

Hint 2

Two equal resistors in parallel give half the resistance: 2 Ω.

Hint 3

Now it is a divider between two equal 2 Ω resistances.

Worked solution — try the problem first

Combine the parallel pair first. Two equal resistors in parallel give half the value:

$R_{eq} = \frac{4 \times 4}{4 + 4} = 2\ \Omega$

Now it is a two-resistor divider — 2 Ω in series with 2 Ω:

$V_A = 10 \times \frac{2}{2 + 2} = 10 \times 0.5 = 5.0\ \text{V}$

$\boxed{V_A = 5.0\ \text{V}}$

Check by KCL at node A. Current in through the source resistor must equal current out through both branches:

$\frac{10 - 5}{2} = \frac{5}{4} + \frac{5}{4} \quad \Longrightarrow \quad 2.5 = 1.25 + 1.25 \ \checkmark$

Concepts:Node analysisParallel resistanceVoltage dividerKCL

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