Generalized Kirchhoff's Laws

If we assume all voltages and currents in a circuit are sinusoids of the same frequency $\omega$, they can be represented as complex phasors (vectors), e.g.,

$$v(t)=V_m cos(\omega t+\phi)=Re[{\bf V}(t)]=Re[V_{eff} e^{j\... ...;\sqrt{2}\;e^{j\omega t}]=Re[\dot{V}\;\sqrt{2}\;e^{j\omega t}]$$

and the Kirchhoff's laws can now be stated as:

• Current Law (KCL): The vector sum of the currents into a node is zero $\sum_k \dot{I}_k=0$.
• Voltage Law (KVL): The vector sum of the voltages around a loop is zero $\sum_k \dot{V}_k=0$.

Example 1: Consider three sinusoidal voltage sources $v_1(t)=6\sqrt{2}\;\sin(\omega t)$, $v_2(t)=12\sqrt{2}\;\sin(\omega t+\pi/2)$ and $v_3(t)=4\sqrt{2}\;\sin(\omega t-\pi/2)$ in series. According to KVL, the overall voltage will be the algebraic sum of the three:

$$v(t)=v_1(t)+v_2(t)+v_3(t) =6\sqrt{2}\;\sin(\omega t)+12\sqrt{2}\;\sin(\omega t+\pi/2) +4\sqrt{2}\;\sin(\omega t-\pi/2)$$

While the addition of these sinusoidal functions is not easy to carry out (still remember all the trigonometric identities?), it is quite straight forward to find the vector sum if the voltages are represented as phasors:

$$\dot{V} = \dot{V}_1+\dot{V}_2+\dot{V}_3=6\angle 0^\circ+12\angle 90^\circ+ 4\angle -90^\circ$$

$$= 6+j12-j4=6+j8=10 \angle tan^{-1}(8/6) =10\angle 53.1^\circ$$

The resulting voltage is therefore $v(t)=10\sqrt{2}\;sin(\omega t+53.1^\circ)$