Table 1: A few voltage divider schematics and their solutions The first step is recognizing when you have a voltage divider. Most derivations assume a setup like the one shown in the figure to the right, with a source voltage shown as a circle or a battery, R1 and R2 in series, vs applied across both R₁ and R2, and Vout being measured across R2. + Vs Vout = V2 R2 R₁ + R2 RI R₂ Vo Sometimes we don't define Vout as the voltage across one of the resistors, we define it as the voltage from "ground" up to the junction between the resistors. This is important if one end of R2 is connected to a battery v or other supply, not to ground. In this case, Vout = V2 + Vb and we use the voltage divider formula with the voltage drop across vi and v2 (not vb) to figure out v2. Vout -(vs) + Vb R₂ R₁+R2 Sometimes the voltage across R1 and R2 is not given, and we are instead given the voltage at the ends of R1 and R2 with respect to ground, and we want yout with respect to ground (this is the case in most of the homework problems). Then we need to calculate Vs from Vы1 and Vы2. Vout Rz R1+R2 -(Vb1 - Vb₂) + Vb2- Sometimes a supply voltage connects to more than one path from Vs- to Vs+. In the case shown here, you can ignore R₁ and R2 because any current that goes through R1 and R2 comes from the supply and is not diverted from the R3-R4 path. R₁ and R2 make their own voltage divider, but unless R₁/R2 = R3/R4 the voltages across R2 and R4 will be different." On the other hand, if there are resistors in parallel that come to one junction in the middle of the path from Vs- to Vs+, you need to combine the parallel resistances. Here you could use the parallel resistance formulas - or Rs R₁R3 R₁+R3' R R2R4 R2+R4 1 1 1+and 1 and 6=2+4 RS R1 R3 Then use the voltage divider formula to find. R6 R2 R4 VouT The triangle is ground or common, and is defined as 0 V. Vb1- Vs+ R R 2 Vs+ Vs- Vout Vb2 R₂ + R4 (US) + Vout R R6 Vout = V2 = V4 = V = Rs+R6 Note that you can't ignore R1 and R2 in the calculation unless R₁ = R3 and R2 = R4. These exercises form a foundation for solving a variety of problems, including reactive circuit analysis (RLC circuits) and op-amp circuits. ? The symbol or indicates a voltage that you should calculate. All voltages are relative to the same common voltage (you may call it ground). Bubbles/lollipops represent terminals where voltage is applied or measured. An indicates that a resistor is disconnected at that end; this occurs in figure 3. Your answers may be reported with two significant digits, consistent with the given resistances. Resistor values are typically accurate to only 2 significant digits. For each figure, determine the voltage at the point indicated by the probe. Enter your answers in the associated Canvas quiz. Figure 1 9 V Figure 2 +3V In this schematic diagram, it is assumed that there is a power supply or battery between the 9 V and 0 V terminals, and that the unknown voltage is measured with respect to 0 V, i.e. across the 220 resistor. Unless you are told/shown otherwise, assume that Vout is measured with respect to 0 V. In this diagram, we assume that 110 Ω R1 the 3 V and 7 V 4.7 k terminals are (?) supplied by w 2.2k +7V separate www batteries, and R2 220 Ω that the unknown voltage is still measured with respect to 0 V. Figure 3 Volt meters usually don't let any current in, so we assume here that there is no current through the 510 2 resistor. Fig. 5 +6 V OV Figure 4. Here we +7V want to figure out 6.8 kQ what the unknown Vin(t) WX 510 Q 5.1 ΚΩ -3V voltage Vin must be in order to get an output of 6 V. Applying 6 V between R1 and R2 would not affect Vin. Vin ww 330 Ω ? w 470 Ω www ww R1 39 +6 V R2 15 Fig. 6. Here you may +5 V о PI R1 V1 F②5 write the w 2.2 ΚΩ answer in R1 22 w R2 terms of R1, R2 and R3. OV V2 ли R3 **Find V2 -- V1 **