| Branch current methodThe first and most straightforward network 
                    analysis technique is called the Branch Current Method. 
                    In this method, we assume directions of currents in a 
                    network, then write equations describing their relationships 
                    to each other through Kirchhoff's and Ohm's Laws. Once we 
                    have one equation for every unknown current, we can solve 
                    the simultaneous equations and determine all currents, and 
                    therefore all voltage drops in the network.  Let's use this circuit to illustrate the 
                    method:  
                      The first step is to choose a node (junction 
                    of wires) in the circuit to use as a point of reference for 
                    our unknown currents. I'll choose the node joining the right 
                    of R1, the top of R2, and the left of 
                    R3.  
                      At this node, guess which directions the 
                    three wires' currents take, labeling the three currents as I1, 
                    I2, and I3, respectively. Bear in mind 
                    that these directions of current are speculative at this 
                    point. Fortunately, if it turns out that any of our guesses 
                    were wrong, we will know when we mathematically solve for 
                    the currents (any "wrong" current directions will show up as 
                    negative numbers in our solution).  
                      Kirchhoff's Current Law (KCL) tells us that 
                    the algebraic sum of currents entering and exiting a node 
                    must equal zero, so we can relate these three currents (I1, 
                    I2, and I3) to each other in a single 
                    equation. For the sake of convention, I'll denote any 
                    current entering the node as positive in sign, and 
                    any current exiting the node as negative in sign:  
                      The next step is to label all voltage drop 
                    polarities across resistors according to the assumed 
                    directions of the currents. Remember that the "upstream" end 
                    of a resistor will always be negative, and the "downstream" 
                    end of a resistor positive with respect to each other, since 
                    electrons are negatively charged:  
                      The battery polarities, of course, remain as 
                    they were according to their symbology (short end negative, 
                    long end positive). It is okay if the polarity of a 
                    resistor's voltage drop doesn't match with the polarity of 
                    the nearest battery, so long as the resistor voltage 
                    polarity is correctly based on the assumed direction of 
                    current through it. In some cases we may discover that 
                    current will be forced backwards through a battery, 
                    causing this very effect. The important thing to remember 
                    here is to base all your resistor polarities and subsequent 
                    calculations on the directions of current(s) initially 
                    assumed. As stated earlier, if your assumption happens to be 
                    incorrect, it will be apparent once the equations have been 
                    solved (by means of a negative solution). The magnitude of 
                    the solution, however, will still be correct.  Kirchhoff's Voltage Law (KVL) tells us that 
                    the algebraic sum of all voltages in a loop must equal zero, 
                    so we can create more equations with current terms (I1, 
                    I2, and I3) for our simultaneous 
                    equations. To obtain a KVL equation, we must tally voltage 
                    drops in a loop of the circuit, as though we were measuring 
                    with a real voltmeter. I'll choose to trace the left loop of 
                    this circuit first, starting from the upper-left corner and 
                    moving counter-clockwise (the choice of starting points and 
                    directions is arbitrary). The result will look like this:
                     
                        
 
 
                        
 
 
                        
 
 
                      Having completed our trace of the left loop, 
                    we add these voltage indications together for a sum of zero:
                     
                      Of course, we don't yet know what the 
                    voltage is across R1 or R2, so we 
                    can't insert those values into the equation as numerical 
                    figures at this point. However, we do know that all 
                    three voltages must algebraically add to zero, so the 
                    equation is true. We can go a step further and express the 
                    unknown voltages as the product of the corresponding unknown 
                    currents (I1 and I2) and their 
                    respective resistors, following Ohm's Law (E=IR), as well as 
                    eliminate the 0 term:  
                      Since we know what the values of all the 
                    resistors are in ohms, we can just substitute those figures 
                    into the equation to simplify things a bit:  
                      You might be wondering why we went through 
                    all the trouble of manipulating this equation from its 
                    initial form (-28 + ER2 + ER1). After 
                    all, the last two terms are still unknown, so what advantage 
                    is there to expressing them in terms of unknown voltages or 
                    as unknown currents (multiplied by resistances)? The purpose 
                    in doing this is to get the KVL equation expressed using the
                    same unknown variables as the KCL equation, for this 
                    is a necessary requirement for any simultaneous equation 
                    solution method. To solve for three unknown currents (I1, 
                    I2, and I3), we must have three 
                    equations relating these three currents (not 
                    voltages!) together.  Applying the same steps to the right loop of 
                    the circuit (starting at the chosen node and moving 
                    counter-clockwise), we get another KVL equation:  
                        
 
 
                        
 
 
                        
 
 
                        
 
 
                      Knowing now that the voltage across each 
                    resistor can be and should be expressed as the 
                    product of the corresponding current and the (known) 
                    resistance of each resistor, we can re-write the equation as 
                    such:  
                      Now we have a mathematical system of three 
                    equations (one KCL equation and two KVL equations) and three 
                    unknowns:  
                      For some methods of solution (especially any 
                    method involving a calculator), it is helpful to express 
                    each unknown term in each equation, with any constant value 
                    to the right of the equal sign, and with any "unity" terms 
                    expressed with an explicit coefficient of 1. Re-writing the 
                    equations again, we have:  
                      Using whatever solution techniques are 
                    available to us, we should arrive at a solution for the 
                    three unknown current values:  
                      So, I1 is 5 amps, I2 
                    is 4 amps, and I3 is a negative 1 amp. But what 
                    does "negative" current mean? In this case, it means that 
                    our assumed direction for I3 was opposite 
                    of its real direction. Going back to our original 
                    circuit, we can re-draw the current arrow for I3 
                    (and re-draw the polarity of R3's voltage drop to 
                    match):  
                      Notice how current is being pushed backwards 
                    through battery 2 (electrons flowing "up") due to the higher 
                    voltage of battery 1 (whose current is pointed "down" as it 
                    normally would)! Despite the fact that battery B2's 
                    polarity is trying to push electrons down in that branch of 
                    the circuit, electrons are being forced backwards through it 
                    due to the superior voltage of battery B1. Does 
                    this mean that the stronger battery will always "win" and 
                    the weaker battery always get current forced through it 
                    backwards? No! It actually depends on both the batteries' 
                    relative voltages and the resistor values in the 
                    circuit. The only sure way to determine what's going on is 
                    to take the time to mathematically analyze the network.  Now that we know the magnitude of all 
                    currents in this circuit, we can calculate voltage drops 
                    across all resistors with Ohm's Law (E=IR):  
                      Let us now analyze this network using SPICE 
                    to verify our voltage figures. We could analyze current as 
                    well with SPICE, but since that requires the insertion of 
                    extra components into the circuit, and because we know that 
                    if the voltages are all the same and all the resistances are 
                    the same, the currents must all be the same, I'll opt 
                    for the less complex analysis. Here's a re-drawing of our 
                    circuit, complete with node numbers for SPICE to reference:
                     
                      network analysis example
 v1 1 0
 v2 3 0 dc 7
 r1 1 2 4
 r2 2 0 2
 r3 2 3 1
 .dc v1 28 28 1
 .print dc v(1,2) v(2,0) v(2,3)
 .end
 v1            v(1,2)      v(2)        v(2,3)          
2.800E+01     2.000E+01   8.000E+00   1.000E+00 Sure enough, the voltage figures all turn 
                    out to be the same: 20 volts across R1 (nodes 1 
                    and 2), 8 volts across R2 (nodes 2 and 0), and 1 
                    volt across R3 (nodes 2 and 3). Take note of the 
                    signs of all these voltage figures: they're all positive 
                    values! SPICE bases its polarities on the order in which 
                    nodes are listed, the first node being positive and the 
                    second node negative. For example, a figure of positive (+) 
                    20 volts between nodes 1 and 2 means that node 1 is positive 
                    with respect to node 2. If the figure had come out negative 
                    in the SPICE analysis, we would have known that our actual 
                    polarity was "backwards" (node 1 negative with respect to 
                    node 2). Checking the node orders in the SPICE listing, we 
                    can see that the polarities all match what we determined 
                    through the Branch Current method of analysis.  
                      
                      REVIEW: 
                      Steps to follow for the "Branch Current" 
                      method of analysis: 
                      (1) Choose a node and assume directions of 
                      currents. 
                      (2) Write a KCL equation relating currents 
                      at the node. 
                      (3) Label resistor voltage drop polarities 
                      based on assumed currents. 
                      (4) Write KVL equations for each loop of 
                      the circuit, substituting the product IR for E in each 
                      resistor term of the equations. 
                      (5) Solve for unknown branch currents 
                      (simultaneous equations). 
                      (6) If any solution is negative, then the 
                      assumed direction of current for that solution is wrong!
                      
                      (7) Solve for voltage drops across all 
                      resistors (E=IR).  |