[[ECE 2k1 Exam 1 Solutions]] # General Notes - Voltage is constant through nodes with parallel branches, current is constant through nodes with branches in series - Remember that voltage represents *drop in voltage* between the connected nodes. Can set reference point as anything - For an ideal voltage source, $V_{+}-V_{-}=\text{source value}$ ## Current & Voltage Division - When elements are in series (same current flows thru all), $V_{k}=V_{in} \frac{R_{k}}{R_{1}+R_{2}+\dots+R_{n}}$ meaning voltage splits proportionally to resistance - When elements are in parallel (same voltage across all), $I_{k}=I_{in} \frac{\frac{1}{R_{k}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\dots+\frac{1}{R_{n}}}$ where $I_{1}=I_{in} \frac{R_{2}}{R_{1}+R_{2}}$ when $n=2$, meaning current splits inversely to resistance # Nodal Analysis 1. Identify all essential nodes in the circuit. 2. Choose a reference node (ground), usually the most connected node. 3. Assign voltage variables to the remaining nodes with respect to ground. 4. Use known source constraints: - If a node is directly connected to ground through a voltage source, its voltage is known. - If a voltage source is between two non-reference nodes, form a supernode *(see below)*. 5. Write KCL at each non-reference node or supernode: - Sum of currents leaving or entering the node \(= 0\). - Express each resistor current using Ohm’s law: $I = \frac{V_1 - V_2}{R}$ 6. Add any extra equations from voltage sources or controlling relationships for dependent sources. 7. Solve the system of equations for the node voltages. 8. Compute any requested currents or powers using the node voltages. *Treat currents going into node as negative and currents leaving node as positive* ## Supernode - Use a supernode when a voltage source is between two non-reference nodes - We must use a supernode in this case because the equation $I=\frac{V_{1}-V_{2}}{R}$ works great for resistors because current through a resistor is easy to express in terms of node voltages. But for an ideal voltage source, the current through it is usually unknown and cannot be directly written as $\frac{V_{1}-V_{2}}{R},$ since there is no resistance given 1. Group the nodes together 2. Treat them as one larger node region called a supernode 3. Write one KCL equation around the outside of the supernode 4. Add one extra voltage constraint equation from the source - If a voltage source connects node $V_{1}$ and node $V_{2}$ then: - Write KCL for the whole supernode - Write the source relation, such as $V_{1}-V_{2}=V_s$ where $V_{s}$ is source voltage # Mesh Analysis 1. Make sure the circuit is planar so mesh analysis applies. 2. Identify each mesh (smallest independent loop). 3. Assign a mesh current to each mesh, usually all clockwise. 4. Write KVL around each mesh: - Sum of voltage rises and drops \(= 0\). 5. For resistor voltages: - If the resistor is in only one mesh: $V = RI_k$ - If the resistor is shared by two meshes: $V = R(I_k - I_m)$ 6. Handle current sources carefully: - If a current source lies only in one mesh, that mesh current is known immediately. - If a current source is shared by two meshes, form a supermesh *(see below)* and write: - one KVL equation around the supermesh - one constraint equation from the current source 7. Include dependent source relationships if present. 8. Solve for the mesh currents. 9. Compute any requested voltages, currents, or powers from the mesh currents. ## Supermesh - Use a supermesh when a current source lies on a branch shared by two meshes - Normally in mesh analysis, you write KVL around each mesh - works fine when all element voltages can be expressed in terms of mesh current. For an ideal current source, the voltage across it is unknown, so if a current source is on the border between two meshes, you cannot directly write a normal KVL equation through that branch 1. Combine the two meshes into a larger loop called a supermesh 2. Write one KVL equation around the outside of that larger loop 3. Add one extra current constrain equation - If a current source of value $I_{s}$ lies between mesh currents $I_{1}$ and $I_{2}$ then write: - One KVL equation around the outer boundary of the supermesh - One source constraint, such as $I_{1}-I_{2}=I_{s}$ or the negative of that, depending on current directions and source arrow directions # Source Transformations - Source transformation is a way to replace a practical voltage source with an equivalent practical current source, or vice versa, without changing how the rest of the circuit sees that two-terminal part of the network. Basically converting between Thevenin and Norton forms for a small piece of a circuit - A voltage source $V_{s}$ in series with a resistor $R$ becomes a current source $I_{s}$ in parallel with the same resistor $R$ where $I_{s}=\frac{V_{s}}{R}$ - A current source $I_{s}$ in parallel with a resistor $R$ becomes a voltage source $V_{s}$ in series with the same resistor $R$ where $V_{s}=I_{s}R$ - See  [[The Bare Essentials of Electrical Engineering.pdf#page=134|Example 4.5.1]] # Thevenin & Norton Circuits - Thevenin & Norton equivalents are ways to replace a complicated **two-terminal linear circuit** with a much simpler circuit that behaves the same from the outside. If you connect any load to the two terminals, the load sees the same voltage, current, and power either way - Suppose you have a messy circuit and only care about what happens at terminals $a$ and $b$. Instead of analyzing the full circuit every time the load changes, you replace everything except the load with: - Thevenin: $V_{th}$ in series with $R_{th}$ - Norton: $I_{N}$ in parallel with $R_{N}$ - The two forms convert directly, so if you find one form, the other is basically free $R_{th}=R_{N}$ $V_{th}=I_{N}R_{N}$ $I_{N}=\frac{V_{th}}{R_{th}}$ - $V_{th}$ is the open-circuit voltage -> voltage across the output terminals when nothing is connected to them, which is why $V_{th}=V_{oc}$ - $I_{N}$ is the current that flows if you connect the two output terminals directly with a wire, which is why $I_{N}=I_{sc}$ - $R_{N}$ is the equivalent resistance seen at the terminals, and for linear circuits $R_{th}=R_{N}$ ## Thevenin Equivalent 1. Remove the load 2. Leave the terminals open then find the voltage across the terminals. - Use nodal analysis, mesh analysis, source transformations,... 3. Find $R_{th}$ for which there are two main cases - Only independent sources -> turn them all off then find the equivalent resistance seen from the terminals - Voltage source -> short circuit - Current source -> open circuit - Dependent sources are present -> do not turn off dependent sources, instead: 1. Turn off only the independent sources 2. Apply a test voltage $V_{t}$ or test current $I_{t}$ at the terminals 3. Solve for a response 4. Compute $R_{th}=V_{t} /I_{t}$ ## Norton Equivalent 1. Remove the load 2. Short the terminals together and find the short-circuit current $I_{N}=I_{sc}$ 3. Since $R_{N}=R_{th},$ find the resistance the same way as above