Using the values from this experiment and Ohm’s Law (V=IR), calculate the resist
ID: 1604398 • Letter: U
Question
Using the values from this experiment and Ohm’s Law (V=IR), calculate the resistance of the circuit by using the amplitude of the current at resonance and where V is the amplitude of the applied voltage. Is this resistance equal to 10 (Ohms)? If not, what physical experimental uncertainties could have produced the discrepancy? Show your work.
Frequency
(Hz)
Vr(V)
R()
I=Vr/R
(A)
Frequency
(Hz)
Vr(V)
R()
I=Vr/R
(A)
10
.9198
10
.092
110
2.50
10
.250
20
1.62
10
.162
120
2.49
10
.249
30
2.08
10
.208
130
2.47
10
.247
40
2.38
10
.238
140
2.45
10
.245
50
2.57
10
.257
150
2.42
10
.242
60
2.69
10
.269
160
2.39
10
.239
70
2.75
10
.275
170
2.37
10
.237
80
2.81
10
.281
180
2.33
10
.233
90
2.82
10
.282
190
2.30
10
.230
100
2.82
10
.282
200
2.23
10
.223
Frequency
(Hz)
Vr(V)
R()
I=Vr/R
(A)
Frequency
(Hz)
Vr(V)
R()
I=Vr/R
(A)
10
.9198
10
.092
110
2.50
10
.250
20
1.62
10
.162
120
2.49
10
.249
30
2.08
10
.208
130
2.47
10
.247
40
2.38
10
.238
140
2.45
10
.245
50
2.57
10
.257
150
2.42
10
.242
60
2.69
10
.269
160
2.39
10
.239
70
2.75
10
.275
170
2.37
10
.237
80
2.81
10
.281
180
2.33
10
.233
90
2.82
10
.282
190
2.30
10
.230
100
2.82
10
.282
200
2.23
10
.223
Explanation / Answer
YEs, here from ohms law V = iR
and in each column
R = V/i = 10 ohms
fro each of frequency , Resistance = 10 ohms
-------------------
if at all there is any Discrepency,
this shud be due to temperature and
the amount of heat losses to the surrouding
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