Question

3. In the circuit shown below, the current flowing through Rs is to be measured by an ampermeter (a) What is the ampermeter reading if the internal resistance (R) of the ampermeter is zero. (b) Assume the ampermeter internal resistance (Rm) is 100 W, what is the measurement error caused by the resistance of the measuring instrument? Circuit Parameters Vsı = 12 V Ry = 400 Vs2 = 2 V R2 =6002 Rş = 1k R4 = 5000 Rs = 1k2 M NM WWO

Answer 1

Page 1

P. Given yoor 50022 | Kon- <m IKR 12V 12V A Good R3 6oor R3 2v av M R2 current flowing through R5 is measured by ammeter. (i) Amper meter internal resistance is zero yoor soon Ikan IR5 Goor series Series & Ikuh 12V ) 12v 12v Applying nodal analysis at node ①

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6001 & 400r are in Series. So Req = 600+ 400 = 1ks 5000 & Ikh are in series. So Req=0.5+1 = 1.5kr 7 IRS 1 KL 1.5k2 AM & lkn 12V +) 2V V-12 + vino + v, -2 = 0 |К Ik 1.5K (21.-12) (1.5k) + Wi-2 - = 0 1.5k 3V, - 18+ Vi -2 = 0 uvi=20 [vi= 5v

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IRG - - = 2 mA Vo-2 1.5K 5-2 3/2k current flowing through IRS with zero internal desistance of amper meter is ZMA b) Whem Rm= 10out here loow is units given but of resistance is ris. so loon is taken. О Л. АЛ I 1.5kr IkR 00 л) 2 100u, series. Ika AM 12V 12v T © 2v av Req = 1500 +100 = 1.6kur IRS 1.6kur ma M IKUR Zikr Like 12V & av

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Applying nodal analysis at V1-12 + + V-2 - 0 1.6K 1K 1K (21,-12) (1.6) + vi-2 = 0 1.6 3.201 – 1912 +0,-2=0 4.2v1 = 21.2 Vi = 21.2 = 5.0476 V. Tes - vi-2 4.2 = 5.0476-2 = 1.9047 MA. 1.6k 1.6K measurement 28007 caused by desistance of measuring instoument = IR5 - IRS = 1,9047-2 = -0.0952 mA.

Here, first by keeping resistance of ampermeter as zero, current flowing through R5 is calculated. Then by keeping resistance of ampermeter as 100 ohm, current flowing through R5 is calculated. Difference between these two currents is the error caused by resistance of the measuring instrument.