PPT Slide

4.5 Circuit Differential Equations cont.

Example 4-4 cont.

KVL (left mesh): vS - 2 (di1/dt) + di2/dt = 0 (A.)

KVL (right mesh): di1/dt - 3 di2/dt - v = 0 (B.)

Ohm’s Law: v = 0.5 W ´ i2 (C.)

Differentiate (C.) to obtain: di2/dt = 2 dv/dt (D.)

Substituting (D.) into (B.) yields: di1/dt = 6 dv/dt + v (E.)

Substituting (D.) & (E.) into (A.) yields:

dv/dt + 0.2 v = 0.1 vS