Question

Calculate the equivalent inductance \(L_{\mathrm{eq}}\) of two ideal inductors, \(L_{1}\) and \(L_{2},\) connected in parallel in a circuit. Assume that their mutual inductance is negligible. [Hint: Imagine replacing the two inductors with a single equivalent inductor \(L_{\mathrm{eq}} .\) How is the emf in the parallel equivalent related to the emfs in the two inductors? What about the currents? \(]\)

Short answer

Answer: The equivalent inductance of two ideal inductors connected in parallel is given by the formula: \(L_{\mathrm{eq}} = \frac{L_{1}L_{2}}{L_{1} + L_{2}}\).

Step-by-step solution

Define the relationships for emf and currents in parallel inductors

For inductors in parallel, the emf across each inductor is the same, and currents are inversely proportional to their inductances. Mathematically, we can write these relationships as: 1. \(emf_{L_{1}} = emf_{L_{2}} = emf_{L_{\mathrm{eq}}}\) 2. \(I_{L_{1}}L_{1} = I_{L_{2}}L_{2}\)

Apply Kirchhoff's current law for parallel components

According to Kirchhoff's current law, the total current entering the parallel combination must be equal to the sum of the currents through each inductor: \(I_{\mathrm{total}} = I_{L_{1}} + I_{L_{2}}\)

Use the current relationship to write an equation for the total current

From step 1, we can write the current in each inductor in terms of the other: \(I_{L_{1}} = \frac{I_{L_{2}}L_{2}}{L_{1}}\) Substituting this expression into the equation for the total current from step 2: \(I_{\mathrm{total}} = \frac{I_{L_{2}}L_{2}}{L_{1}} + I_{L_{2}}\)

Define the relationship of equivalent inductance and total current

Since the emf across each inductor is the same, the current through the equivalent inductor is the same as the total current: \(I_{\mathrm{total}} = I_{L_{\mathrm{eq}}}\)

Substitute for \(I_{L_{\mathrm{eq}}}\) in terms of \(I_{L_{2}}\) and inductances

Using the second relationship from step 1, the current through the equivalent inductor can be written as: \(I_{L_{\mathrm{eq}}} = \frac{emf_{L_{\mathrm{eq}}}}{L_{\mathrm{eq}}}\) Since \(emf_{L_{1}} = emf_{L_{2}} = emf_{L_{\mathrm{eq}}}\), we can also write the current through \(L_{2}\) as: \(I_{L_{2}} = \frac{emf_{L_{2}}}{L_{2}}\) Now we can substitute this expression into the equation from step 4: \(I_{\mathrm{total}} = \frac{I_{L_{2}}L_{2}}{L_{1}} + I_{L_{2}} = \frac{emf_{L_{2}}}{L_{2}} \cdot (\frac{L_{2}}{L_{1}} + 1)\)

Equate the expressions for total current and solve for equivalent inductance

Now we can equate the expressions for \(I_{\mathrm{total}}\) from steps 3 and 5: \(\frac{emf_{L_{\mathrm{eq}}}}{L_{\mathrm{eq}}} = \frac{emf_{L_{2}}}{L_{2}} \cdot (\frac{L_{2}}{L_{1}} + 1)\) Since \(emf_{L_{\mathrm{eq}}} = emf_{L_{2}}\), they cancel each other, thus we have: \(\frac{1}{L_{\mathrm{eq}}} = \frac{1}{L_{2}} \cdot (\frac{L_{2}}{L_{1}} + 1)\) Multiply both sides by \(L_{\mathrm{eq}}L_{1}L_{2}\) to solve for \(L_{\mathrm{eq}}\): \(L_{1}L_{2} = L_{\mathrm{eq}}(L_{2}+L_{1})\) Rearrange the equation to find \(L_{\mathrm{eq}}\): \(L_{\mathrm{eq}} = \frac{L_{1}L_{2}}{L_{1} + L_{2}}\) The equivalent inductance of two ideal inductors connected in parallel, with negligible mutual inductance, is given by: \(L_{\mathrm{eq}} = \frac{L_{1}L_{2}}{L_{1} + L_{2}}\)