Lecture #6 ANNOUNCEMENT • Check the Master Schedule posted online for updated information on TA sections & office hours OUTLINE • The Wheatstone bridge circuit • Delta-to-Wye equivalent circuits • Node-voltage circuit analysis method Reading Finish Chapter 3, Chapter 4.1-4.2 EECS40, Fall 2003Lecture 6, Slide 1 Prof. King The Wheatstone Bridge • Circuit used to precisely measure resistances in the range from 1 Ω to 1 MΩ, with ±0.1% accuracy R1 and R2 are resistors with known values Ω R3 is a variable resistor (typically 1 to 11,000 ) Rx is the resistor whose value is to be measured battery R1 R2 + V current detector – R3 Rx variable resistor EECS40, Fall 2003Lecture 6, Slide 2 Prof. King 1 Finding the value of Rx • Adjust R3 until there is no current in the detector R2 Then, Rx = R3 R1 Derivation: KCL => i1 = i3 and i2 = ix KVL => i R = i R and i R = i R R1 R2 3 3 x x 1 1 2 2 i1 i2 + V i R = i R i3 ix 1 3 2 x – R3 Rx R3 Rx = Typically, R / R can be varied 2 1 R R from 0.001 to 1000 in decimal steps 1 2 EECS40, Fall 2003Lecture 6, Slide 3 Prof. King Delta (Pi) and Wye (Tee) Interconnections Delta Pi R R c a c b ab Rb Rb Ra Ra c c a Wye b Tee a b R1 R2 R1 R2 R3 R3 c c EECS40, Fall 2003Lecture 6, Slide 4 Prof. King 2 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits • In order for the Delta interconnection to be equivalent to the Wye interconnection, the resistance between corresponding terminal pairs must be the same Rc ab Rc (Ra + Rb) Rab = = R1 + R2 Ra + Rb + Rc Rb Ra R (R + R ) a c b a b c Rbc = = R2 + R3 Ra + Rb + Rc R1 R2 Rb (Ra + Rc) R = = R + R R3 ca 1 3 Ra + Rb + Rc c EECS40, Fall 2003Lecture 6, Slide 5 Prof. King ∆-Y and Y-∆ Conversion Formulas Delta-to-Wye conversion Wye-to-Delta conversion Rc RbRc abR1R2 + R2R3 + R3R1 R1 = Ra = R + R + R R a b c Rb Ra 1 c RaRc R1R2 + R2R3 + R3R1 R2 = Rb = Ra + Rb + Rc R2 a b RaRb R1R2 + R2R3 + R3R1 R R R3 = 1 2 Rc = Ra + Rb + Rc R3 R3 c EECS40, Fall 2003Lecture 6, Slide 6 Prof. King 3 Resistive Circuits: Summary • Equivalent resistance of k resistors in series: Σk Req = Ri = R1 + R2 + ••• + Rk i=1 • Equivalent resistance of k resistors in parallel: 1 k 1 1 1 1 = Σ = + + ••• + i=1 Req Ri R1 R2 Rk • Voltage divided between 2 series resistors: I + R 1 v1 R1 v + – v = v s − 1 s R2 R1 + R2 • Current divided between 2 parallel resistors: i i R2 1 2 iS R R i1 = is 1 2 R1 + R2 EECS40, Fall 2003Lecture 6, Slide 7 Prof. King Node-Voltage Circuit Analysis Method 1. Choose a reference node (“ground”) Look for the one with the most connections! 2. Define unknown node voltages those which are not fixed by voltage sources 3. Write KCL at each unknown node, expressing current in terms of the node voltages (using the I-V relationships of branch elements) Special cases: floating voltage sources 4. Solve the set of independent equations N equations for N unknown node voltages EECS40, Fall 2003Lecture 6, Slide 8 Prof. King 4 Nodal Analysis: Example #1 R R1 3 + V I - 1 R2 R4 S 1. Choose a reference node. 2. Define the node voltages (except reference node and the one set by the voltage source). 3. Apply KCL at the nodes with unknown voltage. 4. Solve for unknown node voltages. EECS40, Fall 2003Lecture 6, Slide 9 Prof. King Nodal Analysis: Example #2 R1 R Va 5 R3 I V R 1 V 1 2 R4 2 EECS40, Fall 2003Lecture 6, Slide 10 Prof. King 5 Nodal Analysis w/ “Floating Voltage Source” A “floating” voltage source is one for which neither side is connected to the reference node, e.g. VLL in the circuit below: VLL Va Vb - + I1 R2 R4 I2 Problem: We cannot write KCL at nodes a or b because there is no way to express the current through the voltage source in terms of Va-Vb. Solution: Define a “supernode” – that chunk of the circuit containing nodes a and b. Express KCL for this supernode. EECS40, Fall 2003Lecture 6, Slide 11 Prof. King Nodal Analysis: Example #3 supernode VLL Va Vb - + I1 R2 R4 I2 Eq’n 1: KCL at supernode Eq’n 2: Property of voltage source: EECS40, Fall 2003Lecture 6, Slide 12 Prof. King 6.