Multiloop Circuits — Question 8

Section 27-6 · Physics pages 4426–4427

Question 8 Resistance matching eliminates a bridge branch without changing terminal current A bridge has R1 from A to M, R2 from M to B, R3 from A to N, R4 from N to B, and any positive Rb from M to N. A voltage E is applied from B to A. R1 R2 R3 R4 Rb A M N B Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Derive the resistance-ratio condition for zero current through Rb. (b) At balance, derive the equivalent resistance seen by the source and show why it is independent of Rb.
Show solutionHide solution
Question 8 - Solution Resistance matching eliminates a bridge branch without changing terminal current R1 R2 R3 R4 Rb A M N B (a) Method and calculation VM −VB = E R2 R1 + R2 , VN −VB = E R4 R3 + R4 . VM = VN ⇐⇒R2R3 = R1R4 ⇐⇒R1/R2 = R3/R4. (b) Interpretation and further calculation Ib = (VM −VN)/Rb = 0, Is = E R1 + R2 + E R3 + R4 . Req = (R1 + R2)(R3 + R4) R1 + R2 + R3 + R4 . Physical check At balance, varying a finite positive bridge resistance changes neither node potential nor source current.

Question 8

Resistance matching eliminates a bridge branch without changing terminal current

A bridge has R1 from A to M, R2 from M to B, R3 from A to N, R4 from N to B, and any positive Rb from M to N. A voltage E is applied from B to A.

Diagram shown in the original SVG worksheet.

  1. Derive the resistance-ratio condition for zero current through Rb.

  2. At balance, derive the equivalent resistance seen by the source and show why it is independent of Rb.

Question 8 - Solution

Resistance matching eliminates a bridge branch without changing terminal current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

VM−VB=ER2R1+R2,VN−VB=ER4R3+R4.V_M-V_B=E\frac{R_2}{R_1+R_2},\quad V_N-V_B=E\frac{R_4}{R_3+R_4}.VM=VN⇔R2R3=R1R4⇔R1/R2=R3/R4.V_M=V_N\Longleftrightarrow R_2R_3=R_1R_4\Longleftrightarrow R_1/R_2=R_3/R_4.

(b) Interpretation and further calculation

Ib=(VM−VN)/Rb=0,Is=ER1+R2+ER3+R4.I_b=(V_M-V_N)/R_b=0,\quad I_s=\frac E{R_1+R_2}+\frac E{R_3+R_4}.Req=(R1+R2)(R3+R4)R1+R2+R3+R4.R_{\mathrm{eq}}=\frac{(R_1+R_2)(R_3+R_4)}{R_1+R_2+R_3+R_4}.

Physical check

At balance, varying a finite positive bridge resistance changes neither node potential nor source current.