A 21 V source is applied to a network where R1 = 5 Ω is in series with a parallel branch consisting of R2 = 35 Ω and R3 = 14 Ω. What is the current through resistor R2 under these conditions?

Introduction / Context:
This numerical problem checks series–parallel reduction and the use of Ohm's law to find a branch current. The network is a classic case: one resistor in series with a parallel pair. We first find the equivalent resistance seen by the source, then the voltage across the parallel branch, and finally the individual branch current through R2.

Given Data / Assumptions:

• Source voltage Vs = 21 V (DC).
• R1 = 5 Ω in series with the parallel combination of R2 and R3.
• R2 = 35 Ω, R3 = 14 Ω.
• Resistors are ideal; wiring resistance is negligible.

Concept / Approach:
Use series–parallel reduction and Ohm's law. For the parallel branch, compute R23 using 1/R23 = 1/R2 + 1/R3. The total series resistance is R_total = R1 + R23. The series current is I_total = Vs / R_total. The voltage across the parallel branch is V_parallel = Vs − (I_total * R1) (equivalently I_total * R23). Current through R2 is I_R2 = V_parallel / R2.

Step-by-Step Solution:
1) Compute R23: 1/R23 = 1/35 + 1/14 = 0.028571... + 0.071428... = 0.100000Thus R23 = 1 / 0.1 = 10 Ω2) Total series resistance: R_total = R1 + R23 = 5 + 10 = 15 Ω3) Series current: I_total = Vs / R_total = 21 / 15 = 1.4 A4) Voltage across parallel branch: V_parallel = I_total * R23 = 1.4 * 10 = 14 V5) Branch current through R2: I_R2 = V_parallel / R2 = 14 / 35 = 0.4 A = 400 mA

Verification / Alternative check:
Compute I through R3: I_R3 = 14/14 = 1 A. Check that I_R2 + I_R3 = 0.4 + 1.0 = 1.4 A, which equals the series current found earlier—consistent by Kirchhoff's Current Law.

Why Other Options Are Wrong:

• 200 mA, 600 mA, 800 mA: These do not satisfy both the branch voltage (14 V) and Ohm's law for R2 = 35 Ω.

Common Pitfalls:

• Mistaking the parallel equivalent (it is 10 Ω, not 49 Ω).
• Using source voltage directly across R2 without accounting for the series R1 drop.

Final Answer:
400 mA