University:
University of Houston
Course:
PHYS 1301 | College Physics I (lecture)
Academic year:

2022
Views:

320

Pages:

13
Author:

gdjekojaawr4

CIRCUIT ANALYSIS USING KVL, KCL, AND OHM’S LAW Analyze This! Usually we know the voltage source values and the resistor values. We will assume that here. How many KVL’s can we write for this circuit? # of KVL’s: 3 How many KCL’s can we write for this circuit? # of KCL’s: 2 Let’s Do It: KVL There are three possible KVL paths. 1 2 3 What can we way about the relationship among these three equations? 1 −𝑣𝑆1 + 𝑣𝑅3 + 𝑣𝑅4 = 0 2 −𝑣𝑅3 − 𝑣𝑅1 + 𝑣𝑆2 + 𝑣𝑅2 = 0 3 −𝑣𝑆1 − 𝑣𝑅1 + 𝑣𝑆2 + 𝑣𝑅2 + 𝑣𝑅4 = 0 Let’s Do It: KVL There are three possible KVL paths. 1 2 2 They are not algebraically −𝑣𝑆1 + 𝑣𝑅3 + 𝑣𝑅4 = 0 independent – we can only use two of them −𝑣𝑅3 − 𝑣𝑅1 + 𝑣𝑆2 + 𝑣𝑅2 = 0 (any two will do). 3 −𝑣𝑆1 − 𝑣𝑅1 + 𝑣𝑆2 + 𝑣𝑅2 + 𝑣𝑅2 = 0 1 Let’s Do It: KCL 1 There are two nodes for KCL. 2 1 2 𝑖1 = 𝑖2 + 𝑖3 𝑖2 + 𝑖3 = 𝑖1 What can we say about these two KCL equations? Let’s Do It: KCL 1 Two nodes have been labeled. 1 2 𝑖1 = 𝑖2 + 𝑖3 𝑖2 + 𝑖3 = 𝑖1 They are algebraically identical. We can only use one of them. Let’s Do It: KCL and KVL We have three equations: 2 KVL’s and 1 KCL. Remember: we assume we know the voltage sources and resistor values. 1 𝑖1 = 𝑖2 + 𝑖3 2 −𝑣𝑆1 + 𝑣𝑅3 + 𝑣𝑅4 = 0 3 −𝑣𝑅3 − 𝑣𝑅1 + 𝑣𝑆2 + 𝑣𝑅2 = 0 But now what? We have too many unknowns to go further – the resistor voltages and the three currents are all unknown. Ohm’s Law!! 𝑣𝑅1 = −𝑖3 𝑅1 𝑣𝑅2 = 𝑖3 𝑅2 𝑣𝑅3 = 𝑖2 𝑅3 𝑣𝑅4 = 𝑖1 𝑅4 We can use Ohm’s Law to eliminate the resistor voltages…watch the signs! Substitute these equations into KVL and get… Let’s Do It: KCL and KVL 1 2 3 𝑖1 = 𝑖2 + 𝑖3 −𝑣𝑆1 + 𝑖2 𝑅3 + 𝑖1 𝑅4 = 0 −𝑖2 𝑅3 + 𝑖3 𝑅1 + 𝑣𝑆2 + 𝑖3 𝑅2 = 0 Now we have three equations in three unknown currents! We can solve for the currents and then use Ohm’s Law to get the resistor voltages. YOU TRY! You Try! 1. 2. 3. 4. Label 3 currents and all resistor voltages. Watch the label rules! Write 2 KVL equations and 1 KCL equation. Write Ohm’s Law for each of the resistor voltage variables. Re-write your equations in terms of three unknown currents. Numerical Values Ra = 1.1 [kW] Rb = 4.7 [kW] Rc = 150 [W] Rd = 2.2 [kW] My Answers - va + Ra = 1.1 [kW] Rb = 4.7 [kW] Rc = 150 [W] Rd = 2.2 [kW] + vc i1 i2 - i3 - vd + - vb + If your currents and voltages are labeled the way I have them, then the circuit variables are… i1 = -0.39497 [mA] i2 = -1.31070 [mA] i3 = -0.91562 [mA] va = 0.4345 [V] vb = 4.3034 [V] vc = -0.1966 [V] vd = -0.8689 [V] If you have labeled things differently, your results will be different.