The Ammeter and the Voltmeter — Question 5

Section 27-7 · Physics pages 4440–4441

Question 5 A voltmeter across the specimen makes the ammeter include a second current A voltmeter of resistance RV=10.0 kilohm is connected across an unknown resistor R. An upstream ideal ammeter reads the sum of specimen and voltmeter currents. The readings are V=4.00 V and IA=5.00 mA. + − E A R V Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Correct the naive ratio V/IA to obtain R. (b) Derive the general fractional underestimate in the naive resistance and identify when this arrangement is accurate.
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Question 5 - Solution A voltmeter across the specimen makes the ammeter include a second current + − E A R V (a) Method and calculation IV = V/RV = .400 mA, IR = IA −IV = 4.60 mA. R = V IA −V/RV = 20000 23 Ω≃869.6 Ω. (b) Interpretation and further calculation Rnaive = V/IA = R ∥RV , R −Rnaive R = R R + RV . R ≪RV ⇒Rnaive ≃R. Physical check The ammeter current belongs to the entire parallel combination, not to the specimen alone.

Question 5

A voltmeter across the specimen makes the ammeter include a second current

A voltmeter of resistance RV=10.0 kilohm is connected across an unknown resistor R. An upstream ideal ammeter reads the sum of specimen and voltmeter currents. The readings are V=4.00 V and IA=5.00 mA.

Diagram shown in the original SVG worksheet.

  1. Correct the naive ratio V/IA to obtain R.

  2. Derive the general fractional underestimate in the naive resistance and identify when this arrangement is accurate.

Question 5 - Solution

A voltmeter across the specimen makes the ammeter include a second current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

IV=V/RV=.400mA,IR=IA−IV=4.60mA.I_V=V/R_V=.400\,\mathrm{mA},\quad I_R=I_A-I_V=4.60\,\mathrm{mA}.R=VIA−V/RV=2000023Ω≃869.6Ω.R=\frac V{I_A-V/R_V}=\frac{20000}{23}\,\Omega\simeq869.6\,\Omega.

(b) Interpretation and further calculation

Rnaive=V/IA=R∥RV,R−RnaiveR=RR+RV.R_{\mathrm{naive}}=V/I_A=R\parallel R_V,\quad\frac{R-R_{\mathrm{naive}}}R=\frac R{R+R_V}.R≪RV⇒Rnaive≃R.R\ll R_V\Rightarrow R_{\mathrm{naive}}\simeq R.

Physical check

The ammeter current belongs to the entire parallel combination, not to the specimen alone.