Question

A circuit containing resistance $R_1$, inductance $L_1$ and capacitance $C_1$ connected in series resonates at the same frequency $f_0$ as another circuit containing $R_2$, $L_2$ and $C_2$ in series. If two circuits are connected in series then the new frequency at resonance is

• A. $\frac{3}{4} f_r$
• B. $\frac{3}{2} f_r$
• C. $2 f_r$
• D. $f_r$

Hint:

If two individual networks independently resonate at the exact same frequency, joining them in a simple series combination will never change that resonant frequency. The inductive reactance increase is perfectly offset by the capacitive reactance increase across the combination.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given two separate series RLC circuits that share an identical resonant frequency $f_0$ (denoted also as $f_r$). We need to determine the resonant frequency of a grand combined circuit when all components from both individual circuits are placed together in a single continuous series link.

Step 2: Key Formula or Approach:
The resonant angular frequency of a series LCR loop is defined by:
$$\omega_r = \frac{1}{\sqrt{LC}}$$ For circuit 1: $\omega_r = \frac{1}{\sqrt{L_1 C_1}}$
For circuit 2: $\omega_r = \frac{1}{\sqrt{L_2 C_2}}$
Since the initial frequencies are identical, we have $L_1 C_1 = L_2 C_2$.
When connected together in series, inductors combine as $L_{\text{eq}} = L_1 + L_2$ and capacitors combine as $C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}$.

Step 3: Detailed Explanation:
Let's find the product of the equivalent inductance and capacitance for the combined series assembly:
$$L_{\text{eq}} C_{\text{eq}} = (L_1 + L_2) \cdot \left(\frac{C_1 C_2}{C_1 + C_2}\right) = \frac{L_1 C_1 C_2 + L_2 C_1 C_2}{C_1 + C_2}$$ Since $L_1 C_1 = L_2 C_2$, we can substitute $L_2 C_2$ in place of $L_1 C_1$ in the first term of the numerator:
$$L_{\text{eq}} C_{\text{eq}} = \frac{(L_2 C_2) C_2 + L_2 C_1 C_2}{C_1 + C_2} = \frac{L_2 C_2 (C_2 + C_1)}{C_1 + C_2}$$ The $(C_1 + C_2)$ terms cancel out from the numerator and denominator:
$$L_{\text{eq}} C_{\text{eq}} = L_2 C_2 = L_1 C_1$$ Now, substitute this equivalent product back into the resonance formula:
$$\omega_{\text{new}} = \frac{1}{\sqrt{L_{\text{eq}} C_{\text{eq}}}} = \frac{1}{\sqrt{L_2 C_2}} = \omega_r$$ Since $\omega = 2\pi f$, the frequency remains entirely unaltered: $f_{\text{new}} = f_r$.

Step 4: Final Answer:
The new resonant frequency is $f_r$, which matches option (D).