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KVL Example Resistor Voltage Divider • Consider a Series of Resistors And

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KVL Example Resistor Voltage Divider • Consider a Series of Resistors And KVL Example Resistor Voltage Divider • Consider a series of resistors and a voltage source • Then using KVL V −V1 −V2 = 0 • Since by Ohm’s law V1 = I1R1 V2 = I1R2 • Then V − I1R1 − I1R2 = V − I1()R1 + R2 = 0 • Thus V 5 I1 = = = 1 mA R1 + R2 2000 + 3000 • i.e. get the resistors in series formula Rtotal = R1 + R2 = 5 KΩ KVL Example Resistor Voltage Divider Continued • What is the voltage across each resistor • Now we can relate V1 and V2 to the applied V • With the substitution V I 1 = R1 + R 2 • Thus V1 VR1 5(2000) V1 = I1R1 = = = 2 V R1 + R2 2000 + 3000 • Similarly for the V2 VR2 5(3000) V1 = I1R2 = = = 3V R1 + R2 2000 + 3000 General Resistor Voltage Divider • Consider a long series of resistors and a voltage source • Then using KVL or series resistance get N V V = I1∑ R j KorKI1 = N j=1 ∑ R j j=1 • The general voltage Vk across resistor Rk is VRk Vk = I1Rk = N ∑ R j j=1 • Note important assumption: current is the same in all Rj Usefulness of Resistor Voltage Divider • A voltage divider can generate several voltages from a fixed source • Common circuits (eg IC’s) have one supply voltage • Use voltage dividers to create other values at low cost/complexity • Eg. Need different supply voltages for many transistors • Eg. Common computer outputs 5V (called TTL) • But modern chips (CMOS) are lower voltage (eg. 2.5 or 1.8V) • Quick interface – use a voltage divider on computer output • Gives desired input to the chip Variable Voltage and Resistor Voltage Divider • If have one fixed and one variable resistor (rheostat) • Changing variable resistor controls out Voltage across rheostat • Simple power supplies use this • Warning: ideally no additional loads can be applied. • Loads are current drawing devices • In practice the load resistance >> the divider output resistor • Best if Rload >100Rk Current Divider: Example of KCL • KCL equivalent of voltage divider is a current divider • Consider a current source with resistors in parallel • At node 1 the KCL laws state: I − I1 − I2 = 0 • Define V1 as the voltage between node 1 & node 0 • Then V1 V1 I1 = KKI2 = R1 R2 • Thus from KCL V1 V1 1 1 I = I1 + I2 = + = V1 + R1 R2 R1 R2 • This produces the parallel resistors formula 1 1 V1 I = V1 + = R1 R2 Rtotal Current Divider Continued • To get the currents through R1 and R2 V1 V1 I1 = KK I 2 = R1 R2 • First get the voltage from the KCL equation −1 −1 1 1 1 V1 = I + = I R1 R2 Rtotal • Solving for I1 1 V1 R1 I1 = = I R1 1 1 + R1 R2 • Similarly solving for I2 1 V1 R2 I2 = = I R2 1 1 + R1 R2 Example of Current Divider • Consider 4KΩ and 2KΩ in parallel to a 3 mA current source • The by the current divider for I1 1 1 V R 4000 I = 1 = I 1 = 0.003 =1 mA 1 R 1 1 1 1 1 + + 4000 2000 R1 R2 • Similarly for I2 1 1 V R 2000 I = 1 = I 2 = 0.003 = 2 mA 2 R 1 1 1 1 2 + + 4000 2000 R1 R2 • Note the smaller resistor = larger current • Checking: the voltage across the resistors V1 = I1R1 = 0.001× 4000 = 4V V1 = I2R2 = 0.002× 2000 = 4V General Current Divider • Using KCL to get the currents into the node n I = ∑ I j j=1 • Getting the voltage from the KCL equation −1 −1 N 1 1 V1 = I ∑ = I j=1 Rj Rtotal • Solving for Rk with it’s the current Ik 1 Rk Ik = I N 1 ∑ j=1 R j Practical Current Divider • Create current dividers for use with current sources • Less common that Voltage dividers as a circuit application • Again any load used must not be significant • Load in this case in series with the output resistor • Load must be very small compared to Rk • Best if load is <<0.01 of Rk General Current Divider using Conductance • Often better with parallel circuits to use conductance • Again the KCL says at the node N I = ∑ I j j=1 • Total conductance is resistors in parallel is N N 1 Gtotal = ∑ G j = ∑ j =1 j =1 R j • The general current divider equation for Ik through resistor Rk IGk Ik = N ∑G j j=1 • conductance calculations useful for parallel resistors • conductance equation is I equivalent of voltage divider equation • Note for resistors in series then conductance is 1 N 1 = ∑ Gtotal j=1 G j Solving Circuits with Equivalent Resistors • Series and parallel Resistor equivalents can solve some circuits • Method, make equivalent resistance to simplify • Go between series and parallel as needed • Produce one final equivalent resistance • Use voltage and current divider equations • Get I & V for each element Example Solving Circuits with equivalent Resistors • Consider circuit with R2, R3 in parallel R1 all in series with R4 • For the R2, R3 side R2+3 = R2 + R3 =1000 + 3000 = 4000 • Now get the parallel equivalent 1 1 1 1 1 3 = + = + = R1||2+3 R1 R2+3 2000 4000 4000 4000 R = = 1333.3 Ω 1||2+3 3 • Adding the series resistance Rtotal = R4 + R1||2+3 =1000 +1333.3 = 2333.3 Ω • Thus current from the source is V 14 Itotal = = = 6 mA Rtotal 2333.3 Example Circuits with equivalent Resistors Continued • Voltage across R4 and parallel section is VR4 = I4 R4 = 1000 ×0.006 = 6 V V1 = V − I4 R4 = 14 − 1000 ×0.006 = 8 V • And the current in the parallel resistors V1 8 I1 = = = 4 mA R1 2000 V1 8 I2 = = = 2 mA R2+3 4000 • Solving for the voltages VR2 = I2 R2 = 0.002 ×1000 = 2 V VR3 = I2 R3 = 0.002 × 3000 = 6 V Advantages & Disadvantages: equivalent Resistors Method Advantages • Simply guided by pattern of circuit • Easy to understand Disadvantages • Can be quite time consuming • Some circuits cannot be solved this way Measuring Small Values: - Wheatstone Bridge • Resistor dividers are set by ratios of resistance • Thus can compare unknown R to a known set of R • Called a Wheatstone Bridge • Left side know resistance R1 and variable resistor R3 • Right side known R2 and unknown Rs • Place a very sensitive meter between the middle nodes • Best is a galvonometer • Voltages balance and no current ig flows if R R 3 = s R1 R2 • If know the R1 R2 R3 very accurately can measure Rs accurately R3 Rs = R2 R1 • Must use very accurate variable resistance Circuit Analysis with Kirchhoff's Laws Circuits (EC 4) • Task of Circuit analysis: • Find the current in and the voltage across every element Four methods used: • Resistor substitution • Mesh analysis (KVL) • Node analysis (KCL) • Superposition (simple circuits) • Computer methods use aspects of these Circuit Definitions • Node: point where several current paths meet: • Branch: a current path connecting only two node • Branch contains 1 or more devices eg resistors • Note: a node may have many branch connections • If 2 nodes are connected by a wire • Then combine them into a single node Linear & Nonlinear Circuit Elements Linear devices Response is linear for the applied Voltage or Current eg Double voltage get twice the current eg devices: resistors, capacitors, inductors (coils) Non-Linear devices Response is non-linear for applied Voltage or Current eg may have different response for different polarity of V eg devices Semiconductor Diodes, iron core inductors Kirchhoff's Laws and complex circuits • Kirchoff's laws provide all the equations for a circuit • But if know the currents then can calculate the voltages • If know the voltages then can calculate the currents • Thus only need to solve for one or the other. • Use the other laws to obtain the missing quantity .
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