# Measurement Techniques DC Circuits Feb. 2009. Measurement Techniques DC Circuits Resistance (R) –Ohms, Ω, KΩ, MΩ Voltage (V) –Volt, AC, DC, mV, KV Current.

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Measurement Techniques DC Circuits Feb. 2009

Measurement Techniques DC Circuits Resistance (R) –Ohms, Ω, KΩ, MΩ Voltage (V) –Volt, AC, DC, mV, KV Current (I) –Amp, mA (milliamps), uA (microamps)

Bread Board Techniques - Series Circuits Resistance Measurement RT ΩRT Ω Series Circuit RT = R1 + R2 + R3 R1 R2 R3 RTRT Given R1= 100, R2= 4.7K, R3=100K Find RT Measurement must be made without power applied or wired to the circuit. Individual components must be removed from the circuit to measure the value accurately.

Breadboard Techniques - Series Circuit Voltage Measurement The voltage supplied by the (12V) voltage source is proportionally distributed across each resistor. The higher the resistor value, the greater the voltage drop Kirchoffs Law – The sum of the voltage drop across each resistor in the circuit will add up to the source voltage Vs 12V R1 R2 R3 V R1 V R3 V R2 Vs Vs = V R1 + V R2 + V R3

Calculating Voltage Drops 1.Determine total resistance R T 2.Using Ohms Law calculate total current I T 3.Using Ohms Law again, calculate the voltage drop across R1, R2, R3 Vs 12V R1 R2 R3 V R1 V R3 V R2 Vs R T = R1 + R2 + R3 I T = Vs / R T ITIT V R3 = I T x R3 V R1 = I T x R1 V R2 = I T x R2

Bread Board Techniques - Series Circuit Current Measurement The meter must be configured for current measurement. The circuit must be “opened” and the meter placed (anywhere) in series. The same current flows from the voltage source, “through” the meter, each resistor, and then back to the source. Vs 12V ITIT ITIT ITIT R1 R2 R3 I R1 I R3 I R2 Vs ITIT ITIT I T = I R1 = I R2 = I R3

Bread Board Techniques – Parallel Circuits Resistance RT ΩRT Ω 1 R1 R2 R3RT Circuit must be removed from the voltage source The total resistance is “less than” the smallest resistor value Avoid finger contact when measuring

Parallel Circuits Calculating Total Resistance RT ΩRT Ω Parallel Circuit RT R1 + R2 + R3 1 1 1 1 = R1 R2 R3RT R1//R2//R3 Where R1 is in parallel with R2 which is in parallel with R3

Product-Over-Sum Method Calculate the parallel resistance of any 2 resistors at a time. First do R 1 //R 2 using the Product-Over-Sum method Then use the R 1/2 resistance in parallel with R 3 R1 R2 R3 RTRT R 1 x R 2 R 1 + R 2 Let R p = R 1 // R 2 R p = R p x R 3 R p + R 3 R T = Now R T = R p // R 3

Parallel Circuits Voltage Measurement The source voltage (Vs) is common to all components in the circuit Vs = V R1 = V R2 = V R3 R1 R2 R3 Vs

Parallel Circuits Current Measurement I 1 + I 2 + I 3 Vs R1R2R3 ITIT I1I1 I2I2 I3I3 I T = I 1 + I 2 + I 3 I 2 + I 3

Parallel Circuits Current Calculations ITIT I1I1 I3I3 I2I2 R1R 2 R3 Vs To measure current the circuit must be broken and the current meter must be placed in series with the component.

Calculating Total Current (I T ) 1.First find total resistance R T R1 150 Ω R2 300 Ω R3 100 Ω Vs 50V R1//R2 = (150 x 300) / (150 + 300) = 100 ohms Using Product-Over-Sum Method R p //R 3 = (100 x 100) / (100 + 100) = 50 ohms Using Reciprocal Method 1/R T = 1/R 1 + 1/R 2 + 1/R 3 = 1/150 + 1/300 + 1/100 = 0.00666 + 0.00333 + 0.01 = 0.020 R T = 1/ 0.020 = 50 ohms Note: when the parallel resistors are equal in value, just divide by the number of R’s 3K//3K = 1.5K 6K//6K//6K = 2K 2. Then use Ohm’s Law to find total current

Calculating Total Current (I T ) 1.First find total resistance R T R1 150 Ω R2 300 Ω R3 100 Ω Vs 50V The power supply must be capable of supplying at least 1 amp of current 2. Then use Ohm’s Law to find total current Total Current I T I T = Vs R T 50 v = -------- = 1 amp 50 Ω

Calculating Branch Currents ITIT I1I1 I3I3 I2I2 R1=150R2=300R3=100 Vs 50 V R T = 50 ohms I T = 1 amp I 1 = Vs / R1 = 50/150 = 0.333333 amps I 2 = Vs / R2 = 50/300 = 0.166666 amps I 3 = Vs / R3 = 50/100 = 0.200000 amps 1.00 amp

Series/Parallel Circuits There are multiple current paths. Resistors may be in series or parallel with other resistors. A node is where three or more paths come together. The total power is the sum of the resistors’ power.

Simple Combo circuit --/\/\/\/\-- Rs Reduce the parallel connection to its series equivalent R2 // R3 = Rs E R I Then reduce the series equivalent to the total resistance as seen by the source RT = R1 + Rs

Kirchoff’s says “what goes out come back”

Reduce & Simplify R1 R3 R2 R4 R1 R3 + R2 R4 R T = R 1,2 // R 3,4

Analysis of a combo circuit Calculate 1.Total current 2.Branch currents 3.IR drops Board Solution 100 200 200 400 12 V

Reduce & Simplify – find R T 300 600 R T = R 1,2 // R 3,4 = 300 // 600 = 200 100 200 200 400 12 V12 V 200 Ώ I T = 12 / 200 = 0.06 amps (60 mA)

Determining Total Resistance R1 R2 R3 RTRT R T = V I T 1 1 1 1 RT = R1 + R2 + R3 RTRT V ITIT

Branch Currents 300 600 100 200 200 400 12 V I T = Ia + Ib = 40mA + 20 mA = 60 mA Branch Currents I a = 12 / 300 = 40 mA I b = 12 / 600 = 20 mA I T Ia Ib

IR Drops (voltage across each resistor) R1 R3 100 200 R2 R4 200 400 12 V 60 mA 40 mA 20 mA VR1 = 40 mA x 100 = 4000 mV = 4V VR2 = 40 mA x 200 = 8000 mV = 8V VR3 = 20 mA x 200 = 4000 mV = 4V VR4 = 20 mA x 400 = 8000 mV = 8V

Bridge Circuit V AB In a bridge circuit the voltage difference between the two parallel branches is used to indicate the potential difference between the two points. V A = R2 x Vs R2 + R1 V B = R4 x Vs R4 + R3 Using the Voltage Divider Formula V AB = V A - V B VAVA VBVB R1 R3 R2 R4 Vs A B

Wheatstone Bridge – null balance detector V OUT = 0 volts A balanced bridge can be used to measure an unknown resistance. The Wheatstone bridge can be used as an “ohmmeter” by comparing the unknown resistance value to a known one.

Conditioning circuit for resistive sensors and transducers R1 V OUT R1Rs Vs AB The bridge is often used as a conditioning circuit to convert the output of a resistive type sensing element into a voltage (mV) V OUT can be used to represent some type of process variable Temperature Thermistor Resistance Temperature Detectors (RTD’s) Pressure Strain Gauge Flow Anemometer

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