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? \(]\)

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. \(I_{L_{1}}L_{1} = I_{L_{2}}L_{2}\)

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: \(I_{\mathrm{total}} = I_{L_{1}} + I_{L_{2}}\)

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: \(I_{\mathrm{total}} = \frac{I_{L_{2}}L_{2}}{L_{1}} + I_{L_{2}}\)

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

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: \(I_{\mathrm{total}} = \frac{I_{L_{2}}L_{2}}{L_{1}} + I_{L_{2}} = \frac{emf_{L_{2}}}{L_{2}} \cdot (\frac{L_{2}}{L_{1}} + 1)\)

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: \(L_{\mathrm{eq}} = \frac{L_{1}L_{2}}{L_{1} + L_{2}}\)