TRIAL “x” (units) “y” (units) k = y/x
1 .041 A 4.0 V
2 .083 A 8.2 V
3 .127 A 12.3 V
4 .162 A. 16.0 V
5 .200 A 21.0 V
How do i do this ? ? ?
: Suppose you have collected the data shown below for the circuit in Figure 1.
Voltage is measured across the
resistor. Current is measured
through the resistor. The
exact value of the resistor is
100 Ω.
Figure 1.
DESCRIBE THE MEASURED QUANTITIES:
“x” Electric Current in Amperes .
“y” Voltage in Volts .
DATA:
TRIAL “x” (units) “y” (units) k = y/x
1 .041 A 4.0 V
2 .083 A 8.2 V
3 .127 A 12.3 V
4 .162 A. 16.0 V
5 .200 A 21.0 V
EQUATION OF DATA (USING FORM y = mx + b): .
ACCEPTED VALUE FOR k: .
ACCURACY: AVERAGE k – ACCEPTED k x 100% = .
ACCEPTED k
PRECISION: HIGHEST VALUE OF k x 100% = .
LOWEST VALUE OF k
.
ACCURACY: AVERAGE k – ACCEPTED k x 100% = .
ACCEPTED k
PRECISION: HIGHEST VALUE OF k x 100% = .
LOWEST VALUE OF k
4 answers
“x”(units)“y” (units)k = y/x
1.041 A 4.0 V
2.083 A 8.2 V
3.127 A 12.3 V
4.162 A.16.0 V
5.200 A 21.0 V
1.041 A 4.0 V
2.083 A 8.2 V
3.127 A 12.3 V
4.162 A.16.0 V
5.200 A 21.0 V
NEVER MIND I CANT Get it to post
@SMD^^
Unless you can help it, no one wants to see that.
So keep your dirty thoughts to yourself please.
Thank you
Monkees Fan
Unless you can help it, no one wants to see that.
So keep your dirty thoughts to yourself please.
Thank you
Monkees Fan