P 3.6-19 Determine the values of v1, v2, i3, v4, v5, and i6 in Figure P 3.6-19.

 

Figure P 3.6-19

 

To start to find all of these we need to find the total current which requires the total resistance.  So we start by combining series and parallel combinations of resistors.

 

In the bottom loop, we have a 6 Ω, 6 Ω, and a 4 Ω in series which makes a 16 Ω effective resistance.  This is then in parallel with a 16 Ω resistor making the bottom loop effectively a single 8 Ω resistor.

 

On the left side we have a 20 Ω resistor in parallel with an 80 Ω resistor.

 

 

At the top we have a 10 Ω in series with a 30 Ω  which makes a 40 Ω that is now in parallel with a 10 Ω

 

 

So our circuit now looks like

So we can find total resistance and then total current.

 

 

 

We can calculate v1 and v2 since they are over effective resistances. Notice the polarity of both voltage differences is opposite what the current will provide, so we will get negative values for these.  Also both are effective 8 Ω resistors, so they will be the same voltage drop.

 

 

i3 will be found by current division.  Note i3 is directed opposite to the way i actually flows, so i3 will also be negative.

 

 

i6 is also directed backwards and will be negative.  It also can be found by current division.  Since the effective resistance between the two branches is the same, i6 is simply ½ i3

 

 

Since we know i6, we can now find the voltage drop v5.  The polarity provided for v5 is correct so we will get a positive value.

 

 

Finally to get v4, we need to do a current division to determine how much current flows through the 30 Ω resistor.

 

 

 

 

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This page last updated on January 21, 2019