Two magnetically uncoupled inductive coils have quality factors q 1 and q 2 at a given...
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Two magnetically uncoupled inductive coils have quality factors $q_1$ and $q_2$ at a given operating frequency. Their resistances are $R_1$ and $R_2$.
What will be the effective Q factor at the same operating frequency, if the two coils are connected in series?
Show answer & explanation
The total quality factor of series-connected uncoupled coils is the ratio of the total inductive reactance to the total resistance. Since reactance and resistance both sum linearly in series, the effective Q is the weighted average of the individual Q factors based on their resistances.
Step-by-step Derivation:
Step 1: Define the quality factor for a single coil. The quality factor $q$ is given by $q = \frac{\omega L}{R}$, where $\omega L$ is the inductive reactance $X_L$ and $R$ is the resistance.
Step 2: Express the reactances of the two coils in terms of their Q factors and resistances: $X_{L1} = q_1 R_1$ and $X_{L2} = q_2 R_2$.
Step 3: For two coils connected in series (and magnetically uncoupled), the total inductance $L_{total} = L_1 + L_2$ and the total resistance $R_{total} = R_1 + R_2$.
Step 4: The total inductive reactance is $X_{L,total} = \omega(L_1 + L_2) = \omega L_1 + \omega L_2 = X_{L1} + X_{L2}$.
Step 5: Substitute the expressions from Step 2 into the total reactance: $X_{L,total} = q_1 R_1 + q_2 R_2$.
Step 6: Calculate the effective quality factor $Q_{eff} = \frac{X_{L,total}}{R_{total}} = \frac{q_1 R_1 + q_2 R_2}{R_1 + R_2}$.