Verified Solution • 2025–26 Session moderate

QPages 198-203: Q50: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q50 Solution: A voltmeter consists of a coil connected in series with a…

⚠️ Common Exam Traps & Conceptual Pitfalls

When a voltmeter has a series resistor, the maximum measurable voltage is $V_{max}=I_{ ext{fs}},(R_{ ext{coil}}+R_{ ext{series}})$. Students often compute $V_{max}=I_{ ext{fs}};R_{ ext{coil}}$ and get 0.25 V, missing the 575 Ω resistor. Always sum all series resistances before multiplying by the full‑scale current.

A voltmeter consists of a

25 \ \Omega

coil connected in series with a

575 \ \Omega

resistor. The coil takes 10 mA for full-scale deflection. What maximum potential difference can be measured by this voltmeter?

Step-by-Step Verified Solution

**Answer:**

V = 6

V. **Step-by-Step Explanation:** Coil

= 25\\ \\Omega

, series resistor

= 575\\ \\Omega

, full-scale current

= 10

mA.

V = I(R_{coil} + R_s) = 0.01 \\times 600 = 6

V [citation:2].

All Questions in this Chapter

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