Kirchoff's Circuit Laws

Kirchoff’s Voltage Law §

The net voltage gain in any closed loop is 0.

$$V_{gain}=V_{drop}$$

Kirchoff’s Current Law §

The net current entering and exiting a junction/node is 0.

$$I_{in}=I_{out}$$

Example Problem with Single Battery §

To solve for the currents at each point, we can use KCL and KVL

KCL §

$$\begin{array}{rl}{I}_1& =I_2+I_3\\ I_3& =I_4+I_5\end{array}$$

KVL §

$$V=IR$$ $$\begin{array}{rl}9V-I_1{R}_1-I_2{R}_2=9V-3I_1-2I_2& =0\\ 9V-I_1{R}_1-I_4{R}_4-I_3{R}_3=9V-3I_1-2I_4-3I_3& =0\\ 9V-I_1{R}_1-I_5{R}_5-I_3{R}_3=9V-3I_1-I_5-3I_3& =0\end{array}$$

Answer §

Combining these 5 equations, we can solve for each current.

Example Problem with Double Batteries §

We can use a similar apporach here.

KCL §

$$I_1=I_2+I_3$$

KVL §

$$\begin{array}{rl}9V-I_1-2I_2& =0\\ 9V-I_1-3I_3-4I_3+9V& =0\end{array}$$

Solution §

Solving for the system of equations, we get

$$\begin{array}{rl}{I}_1& =4.3\\ I_2& =2.35\\ I_3& =1.96\end{array}$$