shantelvasquez0110
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PLEASE HURRYYYYYY:
Students had two batteries and two different resistors. During four trials, they build four different circuits and plan to measure the circuit’s current in Amps according to the following table.





Trial Number

Voltage (V)

Resistance (Ω)

Current (A)

1

1.5

200


2

1.5

100


3

3.0

200


4

3.0

100




For which trial would the students measure the smallest current in the circuit?

Respuesta :

Answer: C. Trial 3

Explanation:

Trial 1 and 2 equal 1.5, Trial 3 equals 1 and Trial 4 equals 3. Trial 3 is the smallest current .

oladayoademosu

The trial for which the students would measure the smallest current is the circuit is trial 2 and trial 3.

To know the trial which generates the smallest current, we need to determine the current in each trial.

Since current I = V/R where V = voltage and R = resistance.

For trial 1, V = 1.5 V and R = 200 Ω

So, I = 1.5 V/200 Ω

= 0.0075 A

= 7.5 mA

For trial 2, V = 1.5 V and R = 100 Ω

So, I = 1.5 V/100 Ω

= 0.015 A

= 15 mA

For trial 3, V = 3 V and R = 200 Ω

So, I = 3 V/200 Ω

= 0.015 A

= 15 mA

For trial 4, V = 3 V and R = 100 Ω

So, I = 3 V/100 Ω

= 0.03 A

= 3 mA

Trial 2 and trial 3 both produce a the smallest current of 15 mA.

So, the trial for which the students would measure the smallest current is the circuit is trial 2 and trial 3.

Learn more about current here:

https://brainly.com/question/996480

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