(a) For two inductors in series as in Fig. 6.81(a), show that the current-division principle is v1= L1/L1+L2 vs, v2= L2/L1+L2 vs assuming that the initial conditions are zero. (b) For two inductors in parallel as in Fig. 6.81(b), show that the current-division principle is i1= L2/L1+L2 is,

Question

(a) For two inductors in series as in Fig. 6.81(a), show that the current-division principle is

v_{1}=\frac{L_{1}}{L_{1}+L_{2}} v_{s}, \quad v_{2}=\frac{L_{2}}{L_{1}+L_{2}} v_{s}

assuming that the initial conditions are zero.

(b) For two inductors in parallel as in Fig. 6.81(b), show that the current-division principle is

i_{1}=\frac{L_{2}}{L_{1}+L_{2}} i_{s}, \quad \mathrm{i}_{2}=\frac{L_{1}}{L_{1}+L_{2}} i_{s}

assuming that the initial conditions are zero.

Accepted Answer

(a) [latex]\mathrm{v}_{\mathrm{s}}=\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right) \frac{\mathrm{di}}{\mathrm{dt}}[/latex]

[latex]\frac{\mathrm{di}}{\mathrm{dt}}=\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{L}_{1}+\mathrm{L}_{2}}[/latex]

[latex]\mathrm{v}_{1}=\mathrm{L}_{1} \frac{\mathrm{di}}{\mathrm{dt}}, \mathrm{v}_{2}=\mathrm{L}_{2} \frac{\mathrm{di}}{\mathrm{dt}}[/latex]

[latex]\mathrm{v}_{1}=\frac{\mathrm{L}_{1}}{\mathrm{L}_{1}+\mathrm{L}_{2}} \mathrm{v}_{\mathrm{s}}, \mathrm{v}_{\mathrm{L}}=\frac{\mathrm{L}_{2}}{\mathrm{L}_{1}+\mathrm{L}_{2}} \mathrm{v}_{\mathrm{s}}[/latex]

(b) [latex]\mathrm{v}_{\mathrm{i}}=\mathrm{v}_{2}=\mathrm{L}_{1} \frac{\mathrm{di}_{1}}{\mathrm{dt}}=\mathrm{L}_{2} \frac{\mathrm{di}_{2}}{\mathrm{dt}}[/latex]

[latex]\mathrm{i}_{\mathrm{s}}=\mathrm{i}_{1}+\mathrm{i}_{2}[/latex]

[latex]\frac{\mathrm{di}_{\mathrm{s}}}{\mathrm{dt}}=\frac{\mathrm{di}_{1}}{\mathrm{dt}}+\frac{\mathrm{di}_{2}}{\mathrm{dt}}=\frac{\mathrm{v}}{\mathrm{L}_{1}}+\frac{\mathrm{v}}{\mathrm{L}_{2}}=\mathrm{v} \frac{\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)}{\mathrm{L}_{1} \mathrm{L}_{2}}[/latex]

[latex]\mathrm{i}_{1}=\frac{1}{\mathrm{L}_{1}} \int \mathrm{vdt}=\frac{1}{\mathrm{L}_{1}} \int \frac{\mathrm{L}_{1} \mathrm{L}_{2}}{\mathrm{L}_{1}+\mathrm{L}_{2}} \frac{\mathrm{di}_{\mathrm{s}}}{\mathrm{dt}} \mathrm{dt}=\frac{\mathrm{L}_{2}}{\mathrm{L}_{1}+\mathrm{L}_{2}} \mathrm{i}_{\mathrm{s}}[/latex]

[latex]\mathrm{i}_{2}=\frac{1}{\mathrm{L}_{2}} \int \mathrm{vdt}=\frac{1}{\mathrm{L}_{2}} \int \frac{\mathrm{L}_{1} \mathrm{L}_{2}}{\mathrm{L}_{1}+\mathrm{L}_{2}} \frac{\mathrm{di}_{\mathrm{s}}}{\mathrm{dt}} \mathrm{dt}=\frac{\mathrm{L}_{1}}{\mathrm{L}_{1}+\mathrm{L}_{2}} \mathrm{i}_{\mathrm{s}}[/latex]

Question 6.59: (a) For two inductors in series as in Fig. 6.81(a), show tha...

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