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students had two batteries and two different resistors. during four tri…

Question

students had two batteries and two different resistors. during four trials, they build four different circuits and plan measure the circuits current in amps according to the following table:
for which trial would the students measure the smallest current in the circuit?
trial 1
trial 2
trial 3
trial 4

Explanation:

Step1: Recall Ohm's Law

Ohm's Law is \(I=\frac{V}{R}\), where \(I\) is current, \(V\) is voltage, and \(R\) is resistance.

Step2: Calculate current for each trial

  • Trial 1: \(I_1=\frac{V_1}{R_1}=\frac{1.5}{200}= 0.0075\space A\)
  • Trial 2: \(I_2=\frac{V_2}{R_2}=\frac{1.5}{100}=0.015\space A\)
  • Trial 3: \(I_3=\frac{V_3}{R_3}=\frac{3.0}{200}=0.015\space A\)
  • Trial 4: \(I_4=\frac{V_4}{R_4}=\frac{3.0}{100}=0.03\space A\)

Answer:

A. Trial 1