Solution for Question 7 : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper I

Q.7 (a): A voltmeter with an internal resistance of 4750 Ohm is used to measure the voltage across a resistance of 600 Ohm connected in series with a DC series of internal resistance 400 Ohm. What is the error in measurement? 

Sol: Initially, before connecting the voltmeter, the theoretically calculated voltage drop across the resistance under study (i.e. 600 Ohm) is:

Vx = 600 x V / (400 + 600) = 0.6 V

But, when the voltmeter is connected as shown in figure below, the resultant resistance at the voltmeter will be 532.71 Ohms and this leads to change in the voltage drop across the resistance under study. The new voltage drop (Vx') will be:

 Vx' = 532.71 x V / (532.71 + 400) = 0.5711 V

Hence, the % error = 0.6 - 0.5711 = 0.0288 = 2.88%

Q.7 (b): The periodic voltage of the form shown in the Fig. 7 (b) is applied to (i) a true r.m.s. meter (ii) an “average-measuring” – “r.m.s. indicating” meter (iii) a “peak-measuring” – “r.m.s. indicating” meter. Determine the reading of each instrument. 

Sol: The true RMS meter will give the RMS value of the signal.

The Average value will be