Example Solution: Example: Solution Exercises:Example The circuit shown in (a) has been divided into two parts. In (b), the right-hand part has been replaced with an equivalent circuit. The left-hand part of the circuit has not been changed. Determine the value of the current i2. Solution: 1. Determine the value of the resistance R in (b) that makes the circuit in (b) equivalent to the circuit in (a). 10 40 12 20 10 40R⋅=+=Ω+ 2. Find the current i and the voltage v shown in (b). R is in parallel with the 80 Ω resistor. The equivalent resistance, R || 80, is in series with the 16 Ω resistor. Using voltage division gives ()20 80||80 1620 808820 8016 ||80 16 161620 80RvR4 V⋅+= = ⋅= ⋅=⋅++++ Using Ohm’s law 40.2 A20i == Because of the equivalence, the current i and the voltage v shown in (a) are equal to the current i and the voltage v shown in (b).3. Find the current i2 shown from i using current division. 210 100.2 0.04 A 40 mA10 40 50ii=⋅=⋅==+ How can we check our calculations? Let’s apply KVL to the loop shown below We get 212 40 0iiv+−= Notice that this equation involves all three of the values that we calculated. If our values satisfy this equation, it’s likely that they are correct. Substituting the calculated values gives ()()12 0.2 40 0.04 4 2.4 1.6 4 0+−= + −= The calculated values do satisfy the KVL equation and it’s likely that they are correct. Suppose we aren’t satisfied and want to check further. Apply KCL at the top node of the 80-Ω resistor to get 816 80vvi−=+ Substituting the calculated values into this equation gives 84 40.2 0.25 0.05 0.216 80−=+ ⇒ = + The calculated values do satisfy this KCL equation. We can be confident that they are correct.Example: The circuit shown in (a) has been divided into three parts. In (b), the rightmost part has been replaced with an equivalent circuit. The rest of the circuit has not been changed. The circuit is simplified further in (c). Now the middle and rightmost parts have been replaced by a single equivalent resistance. The leftmost part of the circuit is still unchanged. Solution 1. Determine the value of the resistance R1 in (b) that makes the circuit in (b) equivalent to the circuit in (a). 112 6 4 12 6R8⋅=+=+Ω (1) 2. Determine the value of the resistance R2 in (c) that makes the circuit in (c) equivalent to the circuit in (b). First111140||10|| 40||10||8 4 1110.025 0.1 0.125 0.2540 10 8R == = ==++++Ω Then ()()2112 40 ||10 || 12 40 ||10 ||8 16 RR=+ =+ =Ω (2) 3. Find the current i1 and the voltage v1 shown in (c). Using Ohm’s and Kirchhoff’s laws 111112216 2020 16 2.5 40 1 40 2 10 V2vvvvvRR⎛⎞ ⎛⎞−=++=++=+⇒==−⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠ (3) Then using Ohm’s law 112100.625 A16viR−== =− (4) Because circuits (b) and (c) are equivalent, the current i1 and the voltage v1 shown in (b) are equal to the current i1 and the voltage v1 shown in (c). 4. Find the current i2 and the voltage v2 shown in (b). Using voltage division ()()()()()()112140||10||40||10||8 10 4 102.5 V12 40 ||10 ||8 12 412 40 ||10 ||RvvR−−====+++− (5) Then using Ohm’s law 2212.50.3125 A8viR−== =− (6) Because circuits (a) and (b) are equivalent, the current i2 and the voltage v2 shown in (a) are equal to the current i2 and the voltage v2 shown in (b). 5. Find the current i3 shown in (a). Using current division ()231212 0.31250.2083 A12 6 18ii−== =−+ (7)How can we check our calculations? Let’s see if we get the same values using MATLAB. First, collect equations 1-7 in a MATLAB “m-file”: Next, running that m-file gives As expected, theses values agree with our calculated values.Exercises: Problems 3.7-1 and 3.7-2 in Introduction to Electric Circuits are similar to these