Galvanometer To Voltmeter
A galvanometer of resistance 20 (\Omega) shows full-scale deflection at 5 mA. What series resistance turns it into a voltmeter capable of reading up to 10 V?
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Solution
1980 \(\Omega\)
To convert a galvanometer into a voltmeter, a large resistance is connected in series so that the desired full-scale voltage appears across the combination while only the safe full-scale current flows. The required series resistance follows from (V = I_g(G + R)), giving (R = \frac{V}{I_g} - G). Substituting (V = 10) V, (I_g = 5 \times 10^{-3}) A, and (G = 20;\Omega): (R = \frac{10}{5 \times 10^{-3}} - 20 = 2000 - 20 = 1980;\Omega). The value 2000 (\Omega) forgets to subtract the coil resistance (G). The value 1500 (\Omega) results from using a wrong full-scale current. The value 500 (\Omega) comes from misplacing a factor in the milliampere conversion. This applies the NCERT series-resistance method for voltmeter conversion. A plausibility check supports it: the total resistance (R + G = 2000;\Omega) carrying 5 mA yields exactly (2000 \times 0.005 = 10) V, confirming that the meter reaches full scale precisely at the intended voltage. Complementing the ammeter case, an ideal voltmeter is taken to have nearly infinite resistance so that, connected in parallel, it siphons off negligible current and does not perturb the potential difference it is meant to read; the large series resistance is precisely what raises the instrument's resistance to meet that requirement.
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About This Question
- Subject
- physics
- Chapter
- current electricity
- Topic
- galvanometer to voltmeter
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1980 \(\Omega\)
To convert a galvanometer into a voltmeter, a large resistance is connected in series so that the desired full-scale voltage appears across the combination while only the safe full-scale current flows. The required series resistance follows from (V = I_g(G + R)), giving (R = \frac{V}{I_g} - G). Substituting (V = 10) V, (I_g = 5 \times 10^{-3}) A, and (G = 20;\Omega): (R = \frac{10}{5 \times 10^{-3}} - 20 = 2000 - 20 = 1980;\Omega). The value 2000 (\Omega) forgets to subtract the coil resistance (G). The value 1500 (\Omega) results from using a wrong full-scale current. The value 500 (\Omega) comes from misplacing a factor in the milliampere conversion. This applies the NCERT series-resistance method for voltmeter conversion. A plausibility check supports it: the total resistance (R + G = 2000;\Omega) carrying 5 mA yields exactly (2000 \times 0.005 = 10) V, confirming that the meter reaches full scale precisely at the intended voltage. Complementing the ammeter case, an ideal voltmeter is taken to have nearly infinite resistance so that, connected in parallel, it siphons off negligible current and does not perturb the potential difference it is meant to read; the large series resistance is precisely what raises the instrument's resistance to meet that requirement.
This medium difficulty physics question is from the chapter current electricity, covering the topic of galvanometer to voltmeter. It appeared in the 2025 exam.
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