Understanding Inductors in Parallel

When inductors are connected in parallel, the total inductance is calculated as:

\( \frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \ldots + \frac{1}{L_n} \)

This formula is analogous to the one used for calculating the total resistance in a parallel resistor network. The total inductance of a parallel configuration is always less than the smallest individual inductance in the group.

Key Points to Remember:

• In a parallel inductor circuit, the current divides among the inductors based on their inductances.
• Lower inductance paths carry more current, while higher inductance paths carry less current.
• Parallel inductors are commonly used in circuits to achieve a desired inductance value that might not be available as a single component.

Practical Applications:

• Radio frequency (RF) circuits often use parallel inductors for tuning and filtering purposes.
• Power supply designs employ parallel inductors to handle higher currents by distributing the load.

Examples:

Example 1:

Given: \( L_1 = 10 \, \text{mH}, L_2 = 20 \, \text{mH}, L_3 = 30 \, \text{mH} \)

Calculation:

Convert all inductances to Henries: \[ L_1 = 0.01 \, \text{H}, \, L_2 = 0.02 \, \text{H}, \, L_3 = 0.03 \, \text{H} \] Apply the formula: \[ \frac{1}{L_{\text{total}}} = \frac{1}{0.01} + \frac{1}{0.02} + \frac{1}{0.03} = 100 + 50 + 33.33 = 183.33 \] \[ L_{\text{total}} = \frac{1}{183.33} \approx 0.00546 \, \text{H} = 5.46 \, \text{mH} \]

Result: \( L_{\text{total}} = 5.46 \, \text{mH} \)

Example 2:

Given: \( L_1 = 50 \, \mu\text{H}, L_2 = 100 \, \mu\text{H} \)

Calculation:

Convert all inductances to Henries: \[ L_1 = 50 \times 10^{-6} \, \text{H}, \, L_2 = 100 \times 10^{-6} \, \text{H} \] Apply the formula: \[ \frac{1}{L_{\text{total}}} = \frac{1}{50 \times 10^{-6}} + \frac{1}{100 \times 10^{-6}} = 20000 + 10000 = 30000 \] \[ L_{\text{total}} = \frac{1}{30000} \approx 33.33 \, \mu\text{H} \]

Result: \( L_{\text{total}} = 33.33 \, \mu\text{H} \)