Verified Solution • 2025–26 Session moderate

QPages 198-203: Q38: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q38 Solution: Find the current through the resistor shown in the figure.

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently apply the current‑division rule with resistances instead of conductances, using $I_{branch}=I_{total}\frac{R_{other}}{R_{total}}$ which inverts the correct ratio. Remember that for parallel branches $I_{branch}=I_{total}\frac{1/R_{branch}}{1/R_{1}+1/R_{2}+\dots}$, i.e., use conductances (1/Ω) to split the total current correctly.

Find the current through the 10 Ω10 \ \Omega resistor shown in the figure.

Step-by-Step Verified Solution

**Answer:** i=0.5i = 0.5 A. **Step-by-Step Explanation:** Applying Kirchhoff's laws gives i=0.5i = 0.5 A through the 10Omega10\\ \\Omega resistor [citation:2].

All Questions in this Chapter

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