Chapter 6, Problem 59P

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

$v_1=\frac{L_1}{L_1+L_2}{v}_s\text{, }{v}_2=\frac{L_2}{L_1+L_2}{v}_s$

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

$i_1=\frac{L_2}{L_1+L_2}{i}_s\text{, }{i}_2=\frac{L_1}{L_1+L_2}{i}_s$

assuming that the initial conditions are zero.

Figure 6.81

For Prob. 6.59.