Question

If the ac bridge circuit shown below is balanced, the element Z can be a

• A. pure capacitor
• B. pure inductor
• C. R-L series combination
• D. R-L parallel combination

Answer 1

The correct option is A pure capacitor

At balance,

$(R-\frac{j}{\omega C})\left(\frac{R\left(\frac{-j}{\omega C}\right)}{(R-\frac{j}{\omega C})}\right)=RZ$

$R\left(\frac{-j}{\omega C}\right)=R(X+jY)$

[Let Z = X + jY]

Comparing the real and imaginary parts we get,

RX = 0
X = 0

and
$R\left(\frac{-j}{\omega C}\right)=RjY$

$Y=\frac{-1}{\omega C}=\frac{-1}{\omega C}$

$\therefore$ 'Z' is purely capacitive.