Verified Solution • 2025–26 Session moderate

QPages 198-203: Q17: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q17 Solution: The potential difference between the terminals of a battery of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat the battery’s emf as the voltage across the external resistor, forgetting the internal‑resistance drop. Use the terminal voltage $V_{term}=ε- Ir$ (the given 5.8 V) to find the current, then compute $R=V_{term}/I$; do not substitute the emf directly into $V=IR$.

The potential difference between the terminals of a battery of emf 6.0 V and internal resistance 1 Ω1 \ \Omega drops to 5.8 V when connected across an external resistor. Find the resistance of the external resistor.

Step-by-Step Verified Solution

**Answer:** R=29OmegaR = 29\\ \\Omega. **Step-by-Step Explanation:** varepsilon=6\\varepsilon = 6 V, r=1Omegar = 1\\ \\Omega, V=5.8V = 5.8 V. V=varepsilon−irRightarrow5.8=6−i(1)Rightarrowi=0.2V = \\varepsilon - ir \\Rightarrow 5.8 = 6 - i(1) \\Rightarrow i = 0.2 A. R=V/i=5.8/0.2=29OmegaR = V/i = 5.8/0.2 = 29\\ \\Omega [citation:5][citation:12].

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