Question

1. Given that a certain galvanometer has an internal resistance of 254 1 and can handle a maximum current of 250 HA, design an ammeter capable of reading at full- scale deflection (the maximum current without burning up the meter) 10 A and a voltmeter capable of reading at full-scale deflection 10 V. Assume that the space available in the meter for the shunt (for the ammeter) or limiting (for the voltmeter) resistor is no more than 5 mm in length and it mm in cross-sectional area. State specifically what material and dimensions you are using for the both resistors. Your shunt and limiting resistors chosen must be within +5% of your calculated values.

Answer 1

0 Pag. 11 Calvanometer t Ammeter. To convert a galvanometer to ammetes, we have et apply zhant resistance (s) en parelle of it. Let ta hatten S., we know that Imm sa Ja I s I- Ja S4 G I Here ta maximen measurable Curent 210A I a=maxe non allowable current through it =880mA = 250x156A a = Internal resistance of Eelvanometer = 25 ur.

(Page 22 So from s- o Ja I - Ja to asoro 4294 lo - 25oxios S 6.39 yo - Thes the requered shunt resistante. Eralvanomeles Is voltmeter To convert To Gelvaromster t voltmeter we have to apply resistance R sere, with et. In the case the value of the a resistance R t the Here a 254 А Ia 250 NA 250 x 10 v = Maximum voltere to be measured galvanometer lov. by the p. From 254 R= a da 250x100 - Then the Rz 39746 2 requred resistance en

A Now, much that we mnimum have to choose mater at con get maximum resentence R = 397 26 2 resulence 6-35 X los Avarbble lenght be 5 war 540 m Avaibble A = amme nyloom area R=39746 area & let & be To get the siren resistance with lenght I and the sequered resistivity. that B resistance R= pl A 0- Pa RA 3039726 XP 11 i 39746 XTX 10 5x 10-3 un Р = 24.97 So, we have to choose material whore gerist wity en greater than for or equal to p. But here the a problems ten exat. The value of p very high So we can't a get such material conveniently. The resest vity of the metals are of lite onder of 10-8 as we

(Page 24 Let choose 1st the cross sectional area , (which less than (-271-7) tt mn2 So schoose that) Now fram ( 11- we have resit velo RR.A a 387 % hache higher l 39746 x 10 - St 10 X 12x16 7 -3 [ 1-2 45 m2 = 1-24107 8 - ag.4 x 10 Nour. w this our m no mum sos restatuaty - Let us choose Nichrome as the material Nechrome have sesetevety PN - 100x108 Let's choose the targth area Azw 2X10 uur for both ath of them F R=297 46 r (Voltmeter) Le de longle From & is required.

(Pax B) Parchi RA Pw -7 399 UGX 1-2 x 10 to los to 4.869 x 10 3 pre 1, 2.77 min To for the voltmeter, Material - Nichrome (PN = 100 year) length = 4.77mm Arce= (1-2 x 10-7mm} puny a for -3 S 2 C-38 x 10 het length of the sesistant le. Taking area we have from SA PU R= IN / I la= A 6.38 Ko -3 x 12 x 10 x 60 100 x 10 LO 7. C9 710 le 27.65 X10 A for the voltmeter Materal Nichrome. Length de 765410 am Aras 2 1.2410-7 new 2 Amm)
The answers are in the images. Page number is mentioned at right top. Note that expect the value of R and S, the value of dimension, material can vary man to man. Here are three unknown variables, that is resistivity, length and area. So to solve it, we must have to assume a value by try and error method. Here I assume the area and calculated further. Here I choose nichrome as the material as it have very high resistivity value.