Question

The AC bridge shown in the figure is used to measure the impedance Z.

If the bridge is balanced for oscillator frequency f=2 kHz, then the impedance Z will be

• A. $(260+j0)\Omega$
• B. $(0+j200)\Omega$
• C. $(260-j200)\Omega$
• D. $(260+j200)\Omega$

Answer 1

The correct option is A $(260+j0)\Omega$


$Z_{AB}=500\Omega$

$Z_{CD}=Z$

$Z_{BC}=R_{BC}+\frac{1}{j\omega C}$

$=300-\frac{j}{2\pi \times 2\times {10}^3\times 0.398\times {10}^{-6}}$

$Z_{BC}=R_{BC}+\frac{1}{j\omega C}$

$Z_{AD}=300+j2\pi \times 2\times {10}^3\times 15.91\times {10}^{-3}$

$=300+j200\Omega$

At balance,

$Z_{AB}\times Z_{CD}=Z_{BC}\times Z_{AD}$

$Z_{CD}=Z$

$=\frac{Z_{BC}\times Z_{AD}}{Z_{AB}}$

$\Rightarrow Z=\frac{(300-j200)\times (300+j200)}{500}$

$Z=(260+j0)\Omega$