The ac bridge shown in the figure given below is used to

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The ac bridge shown in the figure given below is used to measure the impedance Z.
If the bridge is balanced for oscillator frequency f = 2 kHz, then impedance Z will be

1. (260 + j0)Ω
2. (0 + j200)Ω
3. (260 – j200)Ω
4. (260 + j200)Ω

Correct Option: A

X_L = 2πfL
= 2π × 2000 × 15.91 × 10^–3

X_C =	1
	2πfC

=	1	= 200Ω
	2π × 2000 × 0.398 × 10^-6

Under balanced condition
Z₁ Z₄ = Z₂ Z₃
⇒ R₁ (R₄ + jX₄) = (R₂ + jX₂) (R₃ – jX₃)
⇒ 500(R₄ + jX₄) = (300 + j200) (300 – j200)
⇒ 500(R₄ + jX₄) = 90000 + 40000

⇒ R₄ + jR₄ =	130000	= 260
	500

⇒ Z = 260 + j0
Alternately

At balance	Z_AB	=	Z_BC
	Z_AD		Z_CD

or,	500	=	R + (1/jωc)
	R + jωL		Z

or Z =	(R + jωL)(R + 1/jωc)	= (300 + j2 × 10³ × 3.14 × 2 × 15.91 × 10^-3)
	500

= 300 -	j/2 × 3.14 × 10³ × 2 × 0.398 × 10^-6	= %
	500

=	(300 + j199.82)(300 - j200)
	500

≈ (260 + j0)Ω