Question

which of the following actions will always decrease voltage?
a. decreasing resistance and increasing current
b. decreasing resistance and keeping current constant
c. increasing resistance and keeping current constant
d. increasing resistance and decreasing current

Explanation:

Step1: Recall Ohm's Law

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

Step2: Analyze Option A

If \(R\) decreases and \(I\) increases, the effect on \(V\) is uncertain. For example, if \(R\) changes from \(4\Omega\) to \(2\Omega\) (\(\Delta R=- 2\Omega\)) and \(I\) changes from \(1A\) to \(3A\) (\(\Delta I = + 2A\)), \(V_1=4\times1 = 4V\), \(V_2=2\times3=6V\) (voltage increases).

Step3: Analyze Option B

If \(I\) is constant (\(I = k\)) and \(R\) decreases. Let \(R_1> R_2\), then \(V_1=I\times R_1\) and \(V_2 = I\times R_2\). Since \(I\) is positive, \(V_1>V_2\) (voltage decreases).

Step4: Analyze Option C

If \(I\) is constant (\(I = k\)) and \(R\) increases. Let \(R_1< R_2\), then \(V_1=I\times R_1\) and \(V_2 = I\times R_2\). Since \(I\) is positive, \(V_1

Step5: Analyze Option D

If \(R\) increases and \(I\) decreases, the effect on \(V\) is uncertain. For example, if \(R\) changes from \(2\Omega\) to \(4\Omega\) (\(\Delta R=+ 2\Omega\)) and \(I\) changes from \(3A\) to \(1A\) (\(\Delta I=-2A\)), \(V_1=2\times3 = 6V\), \(V_2=4\times1 = 4V\) (voltage decreases); but if \(R\) changes from \(2\Omega\) to \(3\Omega\) (\(\Delta R = + 1\Omega\)) and \(I\) changes from \(4A\) to \(3A\) (\(\Delta I=-1A\)), \(V_1=2\times4=8V\), \(V_2=3\times3 = 9V\) (voltage increases).

Answer:

B. Decreasing resistance and keeping current constant