BEE2123 TUT4_Bridge Answer

BEE2123 Tutorial 4_DC and AC Bridges

1. An ammeter is connected in series with an unknown resistance Rx and a dc supply. A voltmeter is connected directly across the supply. The ammeter resistance is 10 Ω and the voltmeter sensitivity is 10 kΩ /V. Determine Rx if the ammeter indicates 0.5 A and the voltmeter reading is 500 V. Ans: RX = 990 Ω . Rx A

V

RA = 10 Ω I = 0.5 A

S=10 kΩ /V V = 500 V

V = I R = I ( Rx + RA ) 500 = 0.5 ( Rx + RA ) Rx = 500/0.5 - 10 Ω = 990Ω

2. A Wheatstone Bridge as in figure below has R1 = 1 kΩ , R2 = 100 Ω and R3 is adjustable from 1 kΩ to 5 kΩ . Calculate the maximum and minimum values of R4.

Ans: R4 min = 100 Ω and R4 max = 500 Ω . B 100 Ω

1 kΩ 3V

A

C

1 kΩ - 5 kΩ D R1R4 = R2R3

When R3 = 1000Ω , R2R3 R4 = ---------R1 = 100Ω

100 x 1000 = --------------1000

When R3 = 5000Ω , R2R3 R4 = ---------R1

100 x 5000 = --------------1000

= 500Ω Therefore the minimum value of R4 is 100Ω and its maximum value is 500Ω

3. For a Wheatstone Bridge as in Problem 2, R1 = 1000 Ω , R2 = 4000 Ω , R3 = 100 Ω and R4 = 400 Ω when the bridge is balanced. Internal resistance of the galvanometer is 100 Ω with its measuring sensitivity is 100 mm/µA. Determine the deflection (mm) of the galvanometer caused by an additional of 1 Ω in resistor R4. (Hint: Use Thevenin’s theory due to imbalance in R4) Ans: Deflection (mm) of the galvanometer = 111.60 mm. Current sensitivity of the galvanometer is 100mm/µ A When the resistor at CD arm is 400Ω , the bridge is balance and the galvanometer does not deflect. When the resistor at CD arm increases by 1Ω , the circuit is unbalance and the galvanometer deflects. Voltage and resistance across AC are B 1 kΩ 3V

A

4 kΩ VAC

100 Ω

C D

400 Ω

Thevenin open voltage across AC VAC = V th = VBA – VBC = I1R1 – I2R2 Applying voltage divider, 1000 4000 = 3 ( ------------------) – 3( ------------------ ) 1000 + 100 4000 + 401 =

3 (0.9090909 – 0.9088843)

=

3 x 0.0002066

=

0.0006198

=

0.6198 mV

Thevenin equivalent resistance when the source is shorted, A 1000Ω

100Ω

B

D

4000Ω

401Ω C

Rth = 1000//100 + 4000//401 1000x100 = ------------1000 = -------

4000x401 + ---------------1100

4401

1604000 + ---------11 4401

= 90.909Ω + 364.463Ω = 455.372Ω

Rth=455.372 Ω Vth=0.6198 mV

G

Rg=100 Ω

Current through the galvanometer, I = Vth/ ( Rth + Rg) 0.6198 mV = -------------------------455.372 Ω + 100 Ω = 1.1160 µ A Therefore the deflection of the galvanometer is 1.1160 µ A x 100mm/µ A = 111.60 mm

4. Derive equations for Cx and Rx referring to the figure below.

Ans: RX = R2R3/R1 and CX = R1C3/R2 B R1

R2

A

C C3

Cx

R3

Rx D

Z1Zx = Z2Z3 Z1 Z2 ----- = ----Yx Y3 Z1Y3

=

Z2Yx

Z1 = R1, Z2 = R2, Y3= (1/R3) +(1/(- j/ω C3 )) = (1/R3) – (ω C3/j) = (1/R3) – (ω C3/j) (-j/-j) = (1/R3) + jω C3 Yx = ( 1/Rx) + jω Cx R1 ((1/R3) + jω C3 ) = R2 (( 1/Rx) + jω Cx) (R1/R3) + jω R1C3 = (R2 /Rx) + jω R2Cx Comparing the real and imaginary parts Real part: R1/R3 = R2 /Rx R2 R3 Rx = ------R1 Imaginary part: R1C3 = R2Cx

R1C3 Cx = -------R2

5. An AC Bridge consists of: R1 = 2000 Ω in parallel with C1 = 0.047 µF; R2 = 1000 Ω in series with C2 = 0.47 µF; C3 = 0.5 µF; and Zx is an unknown component. Determine the value for the unknown component if the oscillator frequency is 1 kHz.

Ans: Rx = 40.1Ω and Cx = 0.833µ F

C1

R2 R1

C3

C2

Zx

Z1Zx = Z2Z3 Z2Z3 Zx = ------Z1 Zx = Z2Z3Y1 Y1 = (1/R1) + jω C1 Z2 = R2 – j/ω C2 Z3= - j/ω C3 Zx = (R2 – j/ω C2) (- j/ω C3) ( 1/R1 + jω C1) =

R2 C1 1 j [---- + ---- + j ( ω R2C1 - -------- ) ] (- ---- ) R1 C2 ω R1C2 ω C3

=

R2C1 1 j R2 C1 [ ------- - ------------ ] - -----( ----- + -----) C3 ω 2R1C2C3 ω C3 R1 C2

If R1 = 2000Ω , C1 =0.047µ F, R2 =1000Ω , C2 =0.47µ F, C3 = 0.5µ F and the frequency of the oscillator = 1000Hz, Then, 1000Ω x0.047µ F

1

j

1000

0.047µ F

Zx

=

[-------------------- - -----------------------------------------] - ------------------- ( --------- + -----------) 0.5µ F (2π 1000)2x2000Ω x0.5µ Fx0.47µ F 2π x1000x0.5µ F 2000

0.47µ F Zx = [ 94 – 53.89] – j318.31[0.5 + 0.1] Zx = 40.1 – j191 ∴ Zx = Rx – jXc Where Rx = 40.1Ω , Xc =191 Cx = 1/(2π x1000x191) = 0.833µ F

6. An AC Bridge is balanced on 1000 Hz with these values: C1 = 0.2 µF; R2 = 500 Ω ; R3 = 300 Ω in parallel with C3 = 0.1 µF; and an unknown component in the fourth branch. Determine the value of capacitor, C4 or inductor, L4 which is in series with resistor R4 in the fourth branch.

Ans: Rx = 34.312 Ω and Lx = 28.971 mH C1

R2

R3 C3 Z1Zx = Z2Z3 Z2Z3 Zx = ------Z1 Z2 Zx = ------Y3 Z1 Y3 Z1Zx = Z2 Z1 = - j/ω C1 Z2 = R2 Y 3= (1/R3) + jω C3 -j

1

Zx

( ------ ) (----- + jω C3 ) Zx = R2 ω C1 R3 -j C3 ( -------- + ---- ) Zx = R2 ω R3C1 C1 C3 j ---+ -------R2 C1 ω R3C1 Zx = --------------------- x ----------------------C3 j C3 j ------ - ------------ + ------C1 ω R3C1 C1 ω R3C1 C3 j R2 ( ---- + --------- ) C1 ω R3C1 Zx = -----------------------------------C3 1 ( ------)2 + ( ------- )2 C1 ω R3C1

C3 jR2 ( R2 ---- + --------- ) C1 ω R3C1 --------------------------------C3 1 ( ----- )2 + ( ------- )2 C1 ω R3C1

Zx =

C3 jR2 ( R2 ---- + --------- ) C1 ω R3C1 = ----------------------------------ω 2R32C12 C32 + C12 (----------------------------) ω 2R32C14

Zx =

C3 ω 2R32C14 jR2 ( R2 ---- x ------------------------- ) + ( --------- x C1 ω 2R32C12 C32 + C12 ω R3C1

Zx =

ω 2R2R32C13 C3 ( ------------------------- ) + ω 2R32C12 C32 + C12

ω 2R32C14 -----------------------------) ω 2R32C12 C32 + C12

ω R2 R3 C13 j ( -----------------------------) ω 2R32C12 C32 + C12

∴ Zx = Rx + jXL where

Rx = (

ω 2R2R32C13 C3 --------------------------) , ω 2R32C12 C32 + C12

R2 R3 C13

ω R2 R3 C13 XL = (---------------------------) ω 2R32C12 C32 + C12

Lx = -------------------------ω 2R32C12 C32 + C12 If C1 = 0.2µ F, R2 =500Ω , R3=300Ω , C3=0.1 µ F and the oscillator frequency is 1000Hz.

(2π 1000)2x500x3002x (0.2x10-6)3x(0.1x10-6) Rx = --------------------------------------------------------------------(2π 1000)2x3002x(0.2x10-6)2x(0.1x10-6)2 + (0.2x10-6)2 500x300x(0.2x10-6) 3 Lx = ------------------------------------------------------------------(2π 1000)2x3002x(0.2x10-6)2x(0.1x10-6)2 + (0.2x10-6)2 Rx = 34.3115Ω Lx = 28.9707mH

7.

A Maxwell Bridge as in figure below is used to measure an unknown impedance:

(i)

The components are: R1 = 3138 Ω , C1 = 20 µF, R2 = 50 Ω and R3 = 100 Ω . Determine the unknown impedance in balanced condition. (ii)

Determine the Q factor of the coil if the frequency is 50 Hz.

Ans: i) Rx = 1.593 Ω and Lx = 0.1 H ii) Q = 19.7

C1

R2 R1 Lx R3

i) Z1Zx = Z2Z3

Rx

Zx= Z2Z3Y1 Y1 =1/R1 + jω C1 Z2 = R2 Z3 = R3 1 Zx= R2 R3 (---- + jω C1) R1 R2R3 = ---------- + jω R2 R3 C1 R1 = Rx + jXL = Rx + jω Lx R2R3 Rx = ---------R1 Lx = R2R3C1 If R1= 3138Ω , C1=20µ F, R2=50Ω , R3=100Ω . Then Rx = 50Ω x 100Ω /3138Ω = 1.593Ω Lx = 50Ω x100Ω x20µ F = 0.1 Henry ii) Q Factor of the inductor = XL/R = ω R1C1 = 2π f x 3138Ω x 20µ F = 19.7