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- 2:34 ull 4G I moodle1.du.edu.om An RTD forms one arm of a Wheatstone bridge as shown in the figure. The RTD is used to measure a temperature with the bridge is operated in a balanced mode. The RTD has a resistance of 252 at a temperature of 0°C and a thermal coefficient of resistance a=0.003925C-1. The value of the variable resistance R, must be set to 41.485 Q to balance the bridge circuit. Determine the temperature measured by the RTD where R2=R3=1002 R2 R3 RTD RRTD E, Select one: O a. 4200Find the current l4 for the figure shown below. 100A 40A 12 I1 80A 13 15 10A 14? a. 100A O b. 400A c. 200A O d. 150ASolve 'b' part only hand written plz asap.... I'll upvote plz hand written
- Please help with formula. Thanks. Upvote will be given to you if is goodA Wheatstone bridge is used for measuriug the valuc of change of resistance of a strain gauge wnich forms one of the arms of the bridge. All tbe arms of the bridge including the strain gauge have a resistance of 100 N each. The maximum a!lowable power dissipation from the strain gauge is 250 mW. Determine the value of maximu:n permissible current through the strain gauge and maximum allowable value of bridge supply voltage. Suppose a source of 20 V is available, lind the value of series resistance to be connected between the source and the bridge to limit the input voltage of the bridge to permissible level.In the figure the current in resistance 6 is i = 1.43 A and the resistances are R₁ = R₂ = R3 = 1.68 2, R₂ = 15.1 Q2, R5 = 7.41 Q2, and R6 = 4.57 2. What is the emf of the ideal battery? E www R₁ O a. 25.7V O b. 52.3 V O C. 47.2 V O d. 21.4 V O e. 6.54 V R3 www R₂ www R₁ R₂ 16 R6
- Please assist with this practice problem 1 c,d,e with details on how to do it. Thank you.Determine the nodes/junctions, Current (I), Loops, Voltage drop on each Resistor (R), and Power (P) on each Resistor (R) with the following given and illustration: Use the following given for batteries and resistors (color bands for your value of resistors, use the lower limit). R1 = red red black goldR2 = orange green black silverR3 = white orange black goldR4 = green violet brown goldR5 = violet red black silverR6 = yellow green black silverR7 = brown black black gold V1 (lower left) = 100VV2 (top center) = 150VV3 (lower right) = 200VR2 = 25kQ. Determine the voltage gain V/VIN ? So kz 30 kn a. 7.25 R2 O b. -50.00 10 Kn c. 50.00 40 kn d. -3.12 e. 3.12
- Solve for: a. Id (atleast 4 decimal place) b. Vgs c. Vd (Atleast 3 decimal value) d. Vs a. Id-Blank 1 mA b. Vgs-Blank 2V c. Vd-Blank 3V d. Vs-Blank 4 V Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answer Blank 4 Add your answer c). 1 MQ 1.5V. 12 V 1.210 Ing Vaso oss = 12 mA V₂ =-4VFrom the circuit determine the current through the 10 kilo-ohm resistor at terminals a-b, micro-ampere. a. 238.0952 b. 692.6407 c. 21.6450 d. 123.2412 e. None of the aboveuse these values: R1=50 R2=70