---
title: "AC bridge balance and Schering bridge"
subject: "Instrumentation I"
chapter: 2
key: "semester_4/instrumentation_i:schering-bridge-capacitance"
source: https://www.bctengineer.com/q/semester_4/instrumentation_i/schering-bridge-capacitance
site: BCT Engineering
---

# AC bridge balance and Schering bridge

Instrumentation I · Chapter 2: Theory of measurement

- Asked in: year not given, 8 marks
- Video: [Schering Bridge  | Unknown Capacitance | Electrical Instruments ( EIM ) | Lec - 51](https://www.youtube.com/watch?v=BK5Nioq7Aoc) — Education 4u
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_i/schering-bridge-capacitance
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_i/chapter-2

## Question

Obtain the balance equation for an ac bridge and explain with diagram how Schering bridge can be used to measure unknown capacitance.

## Short answer

General AC bridge balance requires Z1Z3=Z2Z4 with equal magnitudes and angles. The Schering bridge measures unknown capacitance Cx and series loss rx using standard capacitor C2. At balance, Cx = C2R4/R3, rx = R3C4/C2, D=ωC4R4. Used for insulator testing.

## Answer

#### Balance equation of an AC bridge
Four impedance arms Z1, Z2, Z3, Z4 form the bridge; the AC source is across A–C and the detector across B–D.
```text
         A
        / \
      Z1   Z4
      /     \
     B--[D]--D
      \     /
      Z2   Z3
        \ /
         C
```
At balance, the potentials of B and D are equal, so the detector current is zero:
- Voltage drop across Z1 = drop across Z4, and across Z2 = drop across Z3.
- I1 Z1 = I4 Z4 and I1 Z2 = I4 Z3 (same current in each series pair).

Dividing: **Z1 / Z2 = Z4 / Z3**, or **Z1 Z3 = Z2 Z4**.

In polar form: magnitudes |Z1||Z3| = |Z2||Z4| and angles θ1 + θ3 = θ2 + θ4. Both conditions must be met, so two adjustments are needed.

#### Schering bridge for unknown capacitance
```text
          A
        /   \
 Cx,rx(Z1)   R3 (Z4)
      /         \
     B---[D]----D
      \         /
   C2 (Z2)   R4 || C4 (Z3)
        \     /
          C
```
- Arm AB (Z1): unknown capacitor Cx with its series loss resistance rx.
- Arm BC (Z2): standard loss-free capacitor C2.
- Arm CD (Z3): resistor R4 in parallel with variable capacitor C4.
- Arm DA (Z4): variable resistor R3.

Using Z1 Z3 = Z2 Z4:

(rx + 1/jωCx) · R4/(1 + jωC4R4) = R3/(jωC2)

(rx + 1/jωCx) R4 = R3/(jωC2) + R3R4C4/C2

Equating real parts: **rx = R3 C4 / C2**

Equating imaginary parts: **Cx = C2 R4 / R3**

Dissipation factor: D = ωCx·rx = ωC4R4.

#### Procedure
Adjust C4 and R3 alternately until the detector reads null; read R3, R4, C4, C2 and compute Cx. The scale of C4 can be calibrated directly in D. The bridge is widely used for capacitor testing, insulator and cable testing, where very small losses must be found.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What condition gives balance of an AC bridge with arms Z1..Z4 as drawn (Z1 Z3 = Z2 Z4)?
   A. Magnitude condition only
   B. Angle condition only
   C. Both magnitude and phase-angle conditions must be satisfied
   D. Source frequency equal to zero
   Answer: C. Impedances are complex, so both |Z1||Z3| = |Z2||Z4| and theta1+theta3 = theta2+theta4 must hold, requiring two adjustments. (Key point: AC bridge balance)

2. In the Schering bridge, what is in arm CD (Z3)?
   A. R4 in series with C4
   B. Unknown capacitor with rx
   C. R4 in parallel with variable C4
   D. Standard capacitor C2
   Answer: C. Arm CD has R4 in parallel with variable capacitor C4; the unknown Cx with rx is in arm AB. (Key point: Schering arms)

3. At balance, the unknown capacitance in the Schering bridge is:
   A. Cx = C2 R4 / R3
   B. Cx = C2 R3 / R4
   C. Cx = C4 R3 / C2
   D. Cx = R3R4 C2
   Answer: A. Equating imaginary parts gives Cx = C2 R4/R3, and real parts give rx = R3 C4/C2. (Key point: Result for Cx)

4. The dissipation factor of the capacitor measured by the Schering bridge is:
   A. D = C2/C4
   B. D = omega Cx rx = omega C4 R4
   C. D = R3/R4
   D. D = 1/(omega C4 R4)
   Answer: B. D = omega Cx rx = omega C4 R4, so the C4 scale can be calibrated directly in D. (Key point: Dissipation factor)

## Related in Instrumentation I

- [Maxwell bridge balance equation and suitability](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_i%3Amaxwell-bridge-inductance) — Instrumentation I, chapter 2

## Related in other subjects

- [Hydrologic cycle and water balance equation](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Ahydrologic-cycle-water-balance) — Engineering Hydrology, chapter 1, asked 3×
- [Energy balance and Bowen's ratio evaporation equation](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Abowen-ratio-evap) — Engineering Hydrology, chapter 3, asked 2×
- [Capacitance definition and charge equation](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitance-charge-definition) — Basic Electrical Engineering, chapter 4
- [IRR with investment balance diagram](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-balance-diagram-800) — Engineering Economics, chapter 3