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A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. The internal resistance of the instrument is 100 Ω and a full- scale deflection is produced by a DC current of 1 mA. A voltage of 100 V (rms) is applied to the input terminals. The value of R required is?(a) 63.56 Ω(b) 89.83 Ω(c) 89.83 kΩ(d) 141.3 kΩThe question was posed to me during a job interview.This is a very interesting question from Advanced Problems on Indicating Instruments topic in chapter Extension of Instrument Ranges of Electrical Measurements

Answer»

The correct option is (c) 89.83 kΩ

For explanation I would say: VOAverage = 0.636 × \(\sqrt{2}\) Vrms = 0.8993 Vrms

The deflection with AC is 0.8993 times that with DC for the same value of voltage V

SAC = 0.8993 SDC

SDC of a rectifier type instrument is \(\frac{1}{I_{fs}}\) where Ifs is the CURRENT required to produce full scale deflection, Ifs = 1 mA; Rm = 100 Ω; SDC = 10^3 Ω/V

SAC = 0.8993 × 1000 = 899.3 Ω/V. Resistance of MULTIPLIER RS = SAC V – Rm – 2Rd, where Rd is the resistance of diode, for ideal diode Rd = 0

∴ RS = 899.3 × 100 – 100 = 89.83 kΩ.


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