| Methods of Analysis |
| Maintains same current in branch of circuit | ||
| Doesn’t matter how components are connected external to the source | ||
| Direction of current source indicates direction of current flow in branch | ||
| Voltage across current source | ||
| Depends on how other components are connected | ||
| Series circuit | ||
| Current must be same everywhere in circuit | ||
| Current source in a series circuit | ||
| Value of the current for that circuit | ||
| For the circuit shown | ||
| I = 2 mA | ||
| Circuit analysis | ||
| Sometimes convenient to convert between voltage sources and current sources | ||
| To convert from a voltage source to a current source | ||
| Calculate current from E/RS | ||
| RS does not change | |
| Place current source and resistor in parallel |
| Can also convert from a current source to a voltage source | |
| E = I•RS | |
| Place voltage source in series with resistor |
| A load connected to a voltage source or its equivalent current | ||
| Should have same voltage and current for either source | ||
| Although sources are equivalent | ||
| Currents and voltages within sources may differ | ||
| Sources are only equivalent external to terminals | ||
Current Sources in Parallel and Series
| Current sources in parallel | ||
| Simply add together algebraically | ||
| Magnitude and direction of resultant source | ||
| Add currents in one direction | ||
| Subtract currents in opposite direction | ||
Current Sources in Parallel and Series
| Current sources with different values | ||
| Never place in series | ||
| This violates KCL | ||
| For circuits having more than one source | ||
| Use different methods of analysis | ||
| Begin by arbitrarily assigning current directions in each branch | ||
| Label polarities of the voltage drops across all resistors | ||
| Write KVL around all loops | |
| Apply KCL at enough nodes so all branches have been included | |
| Solve resulting equations |
| From KVL: | ||
| 6 - 2I1 + 2I2 - 4 = 0 | ||
| 4 - 2I2 - 4I3 + 2 = 0 | ||
| From KCL: | ||
| I3 = I1 + I2 | ||
| Solve simultaneous equations | ||
| Arbitrarily assign a clockwise current to each interior closed loop (Mesh) | |
| Indicate voltage polarities across all resistors | |
| Write KVL equations |
| Solve resulting simultaneous equations | ||
| Branch currents determined by: | ||
| Algebraically combining loop currents common to branch | ||
| Assign loop currents and voltage polarities | |
| Using KVL: 6 - 2I1 - 2I1 + 2I2 - 4 = 0 | |
| 4 - 2I2 + 2I1 - 4I2 + 2 = 0 | |
| Simplify and solve equations |
| Mutual resistors represent resistors shared between two loops | |
| R12 represents resistor in loop 1 that is shared by loop 1 and loop 2 | |
| Coefficients along principal diagonal will be positive |
| All other coefficients will be negative | |
| Terms will be symmetrical about principal diagonal |
| Convert current sources into equivalent voltage sources | ||
| Assign clockwise currents to each independent closed loop | ||
| Write simultaneous linear equations | ||
| Use format outline or matrix method | ||
| Solve resulting simultaneous equations or matrix equations | |
| Use a calculator or software program to solve |
| Assign a reference node within circuit and indicate node as ground | |
| Convert voltage sources to current sources |
| Assign voltages V1, V2, etc. to remaining nodes | |
| Arbitrarily assign a current direction to each branch where there is no current source |
| Apply KCL to all nodes except reference node | |
| Rewrite each current in terms of voltage | |
| Solve resulting equations for voltages |
| Mutual conductance | ||
| Common to two nodes | ||
| Mutual conductance G23 | ||
| Conductance at Node 2 | ||
| Common to Node 3 | ||
| Conductances at certain nodes are positive | ||
| Mutual conductances are negative | ||
| Equations are written correctly | ||
| Terms will be symmetrical about principal diagonal | ||
| Convert voltage sources into equivalent current sources | |
| Label reference node as ground | |
| Label remaining nodes as V1, V2, etc. |
| Write linear equation for each node or in matrix form | |
| Solve resulting equations for voltages | |
| Method of solution is same as for mesh |
| Resistors connected to a point of Y | ||
| Obtained by finding product of resistors connected to same point in Delta | ||
| Divide by sum of all Delta resistors | ||
| Given a Delta circuit with resistors of 30, 60, and 90 W | ||
| Resulting Y circuit will have resistors of 10, 15, and 30 W | ||
| A Delta resistor is found: | ||
| Taking sum of all two-product combinations of Y resistor values | ||
| Divide by resistance of Y directly opposite resistor being calculated | ||
| For a Y circuit having resistances of 2.4, 3.6, and 4.8 W | ||
| Resulting Delta resistors will be 7.8, 10.4, and 15.6 W | ||