39. Frequency for Potential Difference
Step 1: Circuit Analysis (Voltage Dividers)
The circuit consists of two parallel branches connected to the same source $V$. Let the source voltage be $V$. We treat the bottom wire as ground ($0$ V) and the top wire as potential $V$.
Left Branch (Node A): An inductor $L$ and capacitor $C$ are in series. The voltage at A is determined by the capacitor’s impedance relative to the total branch impedance.
$$ V_A = V \times \frac{Z_C}{Z_L + Z_C} = V \left( \frac{\frac{1}{j\omega C}}{j\omega L + \frac{1}{j\omega C}} \right) = V \left( \frac{1}{1 – \omega^2 LC} \right) $$Right Branch (Node B): A capacitor $C$ and inductor $L$ are in series. The voltage at B is determined by the inductor’s impedance.
$$ V_B = V \times \frac{Z_L}{Z_C + Z_L} = V \left( \frac{j\omega L}{\frac{1}{j\omega C} + j\omega L} \right) = V \left( \frac{-\omega^2 LC}{1 – \omega^2 LC} \right) $$Step 2: Potential Difference V_AB
We calculate the difference $V_A – V_B$:
$$ V_{AB} = V_A – V_B = V \left[ \frac{1}{1 – \omega^2 LC} – \frac{-\omega^2 LC}{1 – \omega^2 LC} \right] $$ $$ V_{AB} = V \left[ \frac{1 + \omega^2 LC}{1 – \omega^2 LC} \right] $$Step 3: Apply Condition
We are given that the peak voltage difference $|V_{AB}|$ is $\eta$ times the source voltage $|V|$.
$$ \left| \frac{1 + \omega^2 LC}{1 – \omega^2 LC} \right| = \eta $$Since $1 + \omega^2 LC$ is always positive, we can write:
$$ \frac{1 + \omega^2 LC}{|1 – \omega^2 LC|} = \eta $$Step 4: Solve for Frequency
There are two cases due to the absolute value modulus:
Case 1: $1 > \omega^2 LC$ (Below Resonance)
$$ \frac{1 + \omega^2 LC}{1 – \omega^2 LC} = \eta $$ $$ 1 + \omega^2 LC = \eta – \eta \omega^2 LC $$ $$ \omega^2 LC (1 + \eta) = \eta – 1 \implies \omega^2 = \frac{1}{LC} \frac{\eta – 1}{\eta + 1} $$Case 2: $\omega^2 LC > 1$ (Above Resonance)
$$ \frac{1 + \omega^2 LC}{-(1 – \omega^2 LC)} = \eta \implies \frac{1 + \omega^2 LC}{\omega^2 LC – 1} = \eta $$ $$ 1 + \omega^2 LC = \eta \omega^2 LC – \eta $$ $$ 1 + \eta = \omega^2 LC (\eta – 1) \implies \omega^2 = \frac{1}{LC} \frac{\eta + 1}{\eta – 1} $$