Physics · Current Electricity
JEE Main 2024 — 1 February, Shift 1 — Question 37
A galvanometer has a resistance of and it allows maximum current of 5 mA . It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance
- Option A:
- Option B:
- Option C:Correct
- Option D:
Answer: C
Step-by-step solution
\begin{array}{*{20}{r}}{R = }&{\frac{V}{{{I_g}}} - {R_g} = \frac{{100}}{{5 \times {{10}^{ - 3}}}} - 50}\\{}&{\; = 20000 - 50}\\{}&{\; = 19950{\rm{\Omega }}}\end{array}
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2024
- Paper
- 1 February, Shift 1
- Subject
- Physics
- Chapter
- Current Electricity
- Topic
- Electrical Measuring Instruments