(Solved): 6.25 (a) Show that the voltage-division rule for two capacitors in series as in Fig. 6.59(a) is v1 ...

6.25 (a) Show that the voltage-division rule for two capacitors in series as in Fig. 6.59(a) is $v_1=\frac{C_2}{C_1+C_2}{v}_s,v_2=\frac{C_1}{C_1+C_2}{v}_s$ assuming that the initial conditions are zero. (a) Figure $6.59$ For Prob. 6.25. (b) For two capacitors in parallel as in Fig. 6.59(b), show that the current-division rule is $i_1=\frac{C_1}{C_1+C_2}{i}_s,i_2=\frac{C_2}{C_1+C_2}{i}_s$ assuming that the initial conditions are zero.