Calculating the Current in a Single-Loop Circuit — Question 5

Section 27-3 · Physics pages 4360–4361

Question 5 A rheostat must cover both required current endpoints A source E=15.0 V has internal resistance r=1.00 ohm. It drives a fixed 4.00 ohm resistor in series with an adjustable rheostat X. The required current range is .500 A to 2.00 A. + − E R X r I Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Find the required minimum and maximum X. (b) Find the largest current possible with a nonnegative X and explain why no rheostat setting can exceed it.
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Question 5 - Solution A rheostat must cover both required current endpoints + − E R X r I (a) Method and calculation I = E/(r + R + X) ⇒X = E/I −r −R. Xmin = 15/2 −5 = 2.50 Ω, Xmax = 15/.5 −5 = 25.0 Ω. (b) Interpretation and further calculation X ≥0 ⇒r + R + X ≥5.00 Ω. I ≤15/5 = 3.00 A. Physical check The rheostat changes resistance; it cannot cancel the positive fixed resistance in a passive loop.

Question 5

A rheostat must cover both required current endpoints

A source E=15.0 V has internal resistance r=1.00 ohm. It drives a fixed 4.00 ohm resistor in series with an adjustable rheostat X. The required current range is .500 A to 2.00 A.

Diagram shown in the original SVG worksheet.

  1. Find the required minimum and maximum X.

  2. Find the largest current possible with a nonnegative X and explain why no rheostat setting can exceed it.

Question 5 - Solution

A rheostat must cover both required current endpoints

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I=E/(r+R+X)⇒X=E/I−r−R.I=E/(r+R+X)\Rightarrow X=E/I-r-R.Xmin=15/2−5=2.50Ω,Xmax=15/.5−5=25.0Ω.X_{\min}=15/2-5=2.50\,\Omega,\quad X_{\max}=15/.5-5=25.0\,\Omega.

(b) Interpretation and further calculation

X≥0⇒r+R+X≥5.00Ω.X\ge0\Rightarrow r+R+X\ge5.00\,\Omega.I≤15/5=3.00A.I\le15/5=3.00\,\mathrm A.

Physical check

The rheostat changes resistance; it cannot cancel the positive fixed resistance in a passive loop.