Ncert Exercises & Solutions

Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.

Hint: The impedance of the circuit depends on the reactance of the inductor and capacitor.

Step 1: We know that inductive reactance,

X_{L} = 2 πfL

$X_{L} = 2 πfL$

and capacitive reactance,

X_{C} = \frac{1}{2 πfC}

$X_{C} = \frac{1}{2 πfC}$

For every high frequency,

(f \to \infty), X_{L} \to \infty and X_{C} \to 0

$(f \to \infty), X_{L} \to \infty and X_{C} \to 0$

Step 2: When the reactance of a circuit is infinite, it will be considered an open circuit.

When the reactance of a circuit is zero, it will be considered short-circuited.

So,

C_{1}, C_{2} \to shorted and L_{1}, L_{2} \to opened .

$C_{1}, C_{2} \to shorted and L_{1}, L_{2} \to opened .$

So, the effective impedance

= R_{eq} = R_{1} + R_{3}

$= R_{eq} = R_{1} + R_{3}$