Introduction to Electrical and Computer Engineering Kirchhoff’s Laws Electrical & Computer Engineering Kirchhoff’s Laws (1 of 15) Kirchhoff’s Laws • Kirchhoff’s Voltage Law (KVL) • Kirchhoff’s Current Law (KCL) – Named after German scientist Gustav Kirchhoff, these laws are used to help analyze more complicated circuits Electrical & Computer Engineering Kirchhoff’s Laws (2 of 15) 1 Kirchhoff’s Voltage Law (KVL) • The algebraic sum of all voltages around any closed path is equal to zero. – A closed path is one that ends at the node it began at, and in which no node is encountered more than once -VS + V1 + V2 + V3 = 0 V1 + V2 + V3 = VS Electrical & Computer Engineering Kirchhoff’s Laws (3 of 15) KVL Example -30 + V1 + V2 = 0 V1 + V2 = 30 Using Ohm’s Law, we find the current coming from the voltage source V1 = 5*I and V2 = 10*I So, 15*I = 30, or I = 2 A. That makes V1 = 10 V and V2 = 20 V Electrical & Computer Engineering Kirchhoff’s Laws (4 of 15) 2 Series Connections Elements are said to be “in series” when only 2 elements are connected to a single node Write KVL: -VS + V1 + V2 + V3 = 0 VS = V1 + V2 + V3 Apply Ohm’s Law: VS = ISR1 + ISR2 + ISR3 = IS (R1 + R2 + R3) VS = ISReq Where Req = R1 + R2 + R3 Each element in series, carries a common current, I (satisfies KCL at each node) Electrical & Computer Engineering Kirchhoff’s Laws (5 of 15) Series Connections (continued) General expression for k resistors connected in series is: ⋯ Note: Req > largest R Electrical & Computer Engineering Kirchhoff’s Laws (6 of 15) 3 Kirchhoff’s Current Law (KCL) • KCL: the algebraic sum of currents entering any node is zero (conservation of charge) – A node is a point of connection of two or more elements. Can also be stated: the sum of currents entering a node is equal to the sum of currents leaving a node Write KCL at the top node: IS = I1 + I2 + I3 Now Apply Ohms’ Law. IS = VS/R1 + VS/R2 + VS/R3 = Vs [1/R1 + 1/R2 + 1/R3] The elements have the same voltage across them (satisfies KVL in each loop) IS = VS/Req Where 1/Req = 1/R1 + 1/R2 + 1/R3 Electrical & Computer Engineering Kirchhoff’s Laws (7 of 15) Parallel Connections • Elements are said to be in parallel if they form a loop containing no other elements – Think of this as elements that have the same top and bottom nodes 1 1 ∑ When you have 3 Resistors (6, 10, and 25 each) in parallel: 1/Req = 1/6 + 1/10 + 1/25 = 0.306 Req = 3.3 Equivalent resistance 3.3 is less than the smallest R (6 ) Note: Req < smallest R Electrical & Computer Engineering Kirchhoff’s Laws (8 of 15) 4 Parallel Connections (continued) Special case with only 2 resistors 1 1 1 2 Resistors (3 and 6 each) in parallel 1/Req = 1/3 + 1/6 = (2/6 + 1/6) = 3/6 = ½ so Req = 2 OR product over sum yields answer more quickly (3*6)/(3+6) = 18/9 = 2 Electrical & Computer Engineering Kirchhoff’s Laws (9 of 15) Parallel Connections (continued) What happens when R1 = R2? Req = (RR)/2R = R/2 (half the original value) 2 Resistors (6 and 6 each) in parallel: 6*6/(6+6) = 36/12 = 3 or more quickly…6/2 = 3 Electrical & Computer Engineering Kirchhoff’s Laws (10 of 15) 5 Series/Parallel Combinations Find Rab Electrical & Computer Engineering Kirchhoff’s Laws (11 of 15) Series/Parallel Combinations • First step, combine the outer two resistors in series into a single equivalent resistance – 1+2=3 • Second step, combine the calculated equivalent resistance in parallel with the 6 resistor – (3*6)/(3+6)=2 • • • • Continue the process with the remaining resistors Third step, 2+4=6 Fourth step, (6*12)/(6+12)=4 Fifth step, 4+6=10=Rab Electrical & Computer Engineering Kirchhoff’s Laws (12 of 15) 6 Series/Parallel Combination Example How much power is the 5 A source supplying? First step, find the equivalent resistance. Electrical & Computer Engineering Kirchhoff’s Laws (13 of 15) Series/Parallel Combination Example 6 10 *64 / 16 64 7.2 12.8 *30 / 30 Electrical & Computer Engineering 16*64 / 16 64 7.2 12.8 20 20*30 / 20 30 12 Kirchhoff’s Laws (14 of 15) 7 Series/Parallel Combination Example 5 12=300W Electrical & Computer Engineering Kirchhoff’s Laws (15 of 15) 8