Nilsson 400 Manual
Our Small Soil Box, which has a 80 ml (80 cm³) volume is compatible with 4-Terminal Resistance Meters such as the, the and the Nilsson 400 Resistance Meters.Miller Soil Boxes can be used to satisfy either the 4-electrode method (ASTM G57 Standard) or the 2-electrode methods (ASTM G187 or AASHTO T-288 Standards).For the 2-electrode method, the two interior pins are removed which changes the multiplication factor from 1 cm (for the 4-electrode method) to either 0.67 cm for the small box or to 0.57 cm for the large box. Please see the following link for details:. Features. Plexiglas body with rounded corners for easy cleaning.
Stainless steel current distribution plates and removable brass pins (potential measurement pins)Critical Dimensions (for 4-electrode method, i.e. The two interior pins are in use): The soil boxes are designed such that the cross-sectional area of the soil (or liquid) sample (A), with the box filled level, divided by the separation between the pins (L) is equal to 1 cm. El discurso filosofico de la modernidad habermas pdf online. Cross-sectional area = 3 cm x 2.4 cm = 7.2 cm². Pin separation = 7.2cm.
A / L = 1 cmSoil Box Example: M.C.Miller Factory Tap Water Measured Resistance= 1.97 kΩResistivity Value= 1.97 kΩ x 1cm= 1,970 Ω.cmHence, MCM Water Conductivity = 507 µS/cm.
Electric circuits 10th edition nilsson solutions manual.1.Electric Circuits 10th Edition Nilsson SOLUTIONS MANUALFull download:ElementsAssea ssment ProblemsAP 2.1a Note that the current ib is in the same circuit branch as the 8 A currentsource; however, ib is defined in the opposite direction of the currentsource. Therefore,ib = −8 ANext, note that the dependent voltage source and the independentvoltage source are in parallel with the same polarity. Therefore, theirvoltages are equal, andvg =ib=4−8= 2 V4−.b To find the power associated with the 8 A source, we need to find thevoltage drop across the source, vi. Note that the two independent sourcesare in parallel, and that the voltages vg and v1 have the same polarities,so these voltages are equal:vi = vg = −2 VUsing the passive sign convention,ps = (8 A)(vi) = (8 A)(−2 V) = −16 WThus the current source generated 16 W of power.2–1.2–2 CHAPTER 2. Circuit ElementsAP 2.2a Note from the circuit that vx = −25 V. To find α note that the twocurrent sources are in the same branch of the circuit but their currentsflow in opposite directions. Thereforeαvx = −15 ASolve the above equation for α and substitute for vx,α =−15 A=−15 A= 0.6 A/Vvx −25 Vb To find the power associated with the voltage source we need to know thecurrent, iv.
Note that this current is in the same branch of the circuit asthe dependent current source and these two currents flow in the samedirection. Therefore, the current iv is the same as the current of thedependent source:iv = αvx = (0.6)(−25) = −15 AUsing the passive sign convention,ps = −(iv)(25 V) = −(−15 A)(25 V) = 375 W.Thus the voltage source dissipates 375 W.AP 2.3a The resistor and the voltage source are in parallel and the resistor voltageand the voltage source have the same polarities. Therefore these twovoltages are the same:vR = vg = 1 kV.i2Problems 2–3Note from the circuit that the current through the resistor is ig = 5 mA.Use Ohm’s law to calculate the value of the resistor:R =vRig1 kV=5 mA= 200 kΩUsing the passive sign convention to calculate the power in the resistor,pR = (vR )(ig ) = (1 kV)(5 mA) = 5 WThe resistor is dissipating 5 W of power.b Note from part (a) the vR = vg and iR = ig. The power delivered by thesource is thuspsource = −vgig so vg = −psourceig=−3 W= 40 V−75 mASince we now have the value of both the voltage and the current for theresistor, we can use Ohm’s law to calculate the resistor value:R =vgig40 V=75 mA= 533.33 ΩThe power absorbed by the resistor must equal the power generated bythe source.
Nilsson Model 400
Thus,pR = −psource = −(−3 W) = 3 Wc Again, note the iR = ig. The power dissipated by the resistor can bedetermined from the resistor’s current:pR = R(iR)2= R(ig )2Solving for ig,=pr=gR480 mW= 0.0016 so i =300 Ωg√0.0016 = 0.04 A = 40 mAThen, since vR = vgvR = RiR = Rig = (300 Ω)(40 mA) = 12 V so vg = 12 VAP 2.4a Note from the circuit that the current through the conductance G is ig,flowing from top to bottom, because the current source and the.gvg g=g2–4 CHAPTER 2.
Nilsson 400 Manual Pdf
Circuit Elementsconductance are in the same branch of the circuit so must have the samecurrent. The voltage drop across the current source is vg, positive at thetop, because the current source and the conductance are also in parallelso must have the same voltage.