1.Calculate the value of the shunt resistor (R,) that must be connected with the galvanometer in order to convert it into an ammeter that can measure a maximum current of Imax=0.75 A. IGMAX RG R = Imax-IGMAX Copper wire: p Copper-1.72x10$ (2.m), D =0.40 mm 2.Calculate the required length (Ls) of a copper wire that has a resistance R, using : R, A L, Ls = where A is the wire's cross sectional area and p is its resistivity. 3.Convert the galvanometer into an ammeter by connecting the shunt resistance R, in v with the galvanometer.

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11th Edition
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Chapter1: Units, Trigonometry. And Vectors
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1- Find Rs 2- Find Ls
1
part I
3 I max
0.75 A
4 RG
5 ohm
5 Igmax
6.
15 mA
7
d(div)
ΙΑ (Α)
8
2
0.10
9.
4
0.20
10
0.30
11
8
0.40
12
10
0.50
13
12
0.60
14
15
0.75
15
2.
Transcribed Image Text:1 part I 3 I max 0.75 A 4 RG 5 ohm 5 Igmax 6. 15 mA 7 d(div) ΙΑ (Α) 8 2 0.10 9. 4 0.20 10 0.30 11 8 0.40 12 10 0.50 13 12 0.60 14 15 0.75 15 2.
1.Calculate the value of the shunt resistor (R,) that must be connected with the galvanometer in order to convert it into an ammeter
that can measure a maximum current of Imax=0.75 A.
IGMAX RG
Imax-IGMAX
Copper wire: p Copper=1.72x10-$ (2.m), D=0.40 mm
2.Calculate the required length (Ls) of a copper wire that has a resistance R, using :
R, A
L,
Ls =
where A
is the wire's cross sectional area and p is its resistivity .
3.Convert the galvanometer into an ammeter by connecting the shunt resistance R, in
with the galvanometer.
Transcribed Image Text:1.Calculate the value of the shunt resistor (R,) that must be connected with the galvanometer in order to convert it into an ammeter that can measure a maximum current of Imax=0.75 A. IGMAX RG Imax-IGMAX Copper wire: p Copper=1.72x10-$ (2.m), D=0.40 mm 2.Calculate the required length (Ls) of a copper wire that has a resistance R, using : R, A L, Ls = where A is the wire's cross sectional area and p is its resistivity . 3.Convert the galvanometer into an ammeter by connecting the shunt resistance R, in with the galvanometer.
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