Q.An ac source is connected across a two-branch ladder network between a left node A and a right node B. The upper branch, from A to B, is a series combination of a resistor $R_1$, then a capacitor $C_1$, then an inductor $L_1$ (in that order). The lower branch, from A to B, is a series combination of an inductor $L_2$, then a resistor $R_2$, then a resistor $R_3$ (in that order). A capacitor $C_2$ is connected vertically as a bridge, joining the junction between $C_1$ and $L_1$ in the upper branch to the junction between $R_2$ and $R_3$ in the lower branch. Draw the effective equivalent circuit of this network at very high frequencies and find the effective impedance.

Concept: what circuit elements do at high frequency

• Capacitor: $X_C=\dfrac{1}{\omega C}\to 0$ $\Rightarrow$ short circuit (a plain wire).
• Inductor: $X_L=\omega L\to\infty$ $\Rightarrow$ open circuit (a break).

Applying it to the network

Replace $C_1$ and $C_2$ by wires, and $L_1$ and $L_2$ by breaks.

1. Upper branch $A\!-\!R_1\!-\!C_1\!-\!M\!-\!L_1\!-\!B$: $C_1$ becomes a wire, so $A$ reaches the mid-node $M$ through $R_1$ only. $L_1$ is now an open break, so $M$ can no longer reach $B$ along the top.
2. Lower branch $A\!-\!L_2\!-\!R_2\!-\!N\!-\!R_3\!-\!B$: $L_2$ is an open break, so $A$ is disconnected from $R_2$ and node $N$ along the bottom-left. The stretch $N\!-\!R_3\!-\!B$ stays intact.
3. Bridge $M\!-\!C_2\!-\!N$: $C_2$ becomes a wire, tying $M$ directly to $N$.

The surviving path

The only continuous route from $A$ to $B$ is …

Method: Simplifying a Reactive Network at a Frequency Limit

This method reduces any complicated network of resistors, inductors, and capacitors to its equivalent circuit at a limiting frequency (very high or very low), turning a hard phasor problem into simple resistor algebra.

Steps

Step 1: Recall how each reactive element behaves at the frequency limit in question.

Capacitor reactance $X_C = 1/(\omega C)$: as $\omega \to \infty$, $X_C \to 0$ (acts like a short, i.e. a plain wire); as $\omega \to 0$, $X_C \to \infty$ (acts like an open, i.e. a break). Inductor reactance $X_L = \omega L$: as $\omega \to \infty$, $X_L \to \infty$ (acts like an open); as $\omega \to 0$, $X_L \to 0$ (acts like a short). These two behave in exactly opposite ways to each other at any given limit.

Step 2: Redraw the network, replacing every reactive element with its limiting behaviour.

Mechanically go component by component: replace every capacitor with a plain connecting wire (or a break, depending on the limit) and every inductor with a break (or a wire). What remains is a purely resistive network — no phasors or complex impedance needed anymore.

Step 3: Trace which resistors actually carry current in the simplified network. …