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- clear answer and correct or will dislikeFor given value of the circuit, Vsupply = 120 V, and with R₁= 2002, R₂=30.0N, R3=75N, R4=30N, R5= 2002, R6-400 and R7=12002? What is the current flow in resistor R4? 1.2 Amp. 1.0 Amp. 0.67 Amp. 2 Amp. V1 -120 V R1 m 2002 R3 750 R7 12002 R6 4002 R2 2300 R4 3002 R5 2005. For the given circuit below, provide the following unknowns: RS RI = 1.2k2 R2 = 2.2k2 R3 R3 = 3.3k2 R4 = 4.4k2 R5 = 5.5k2 R6 = 6.6k2 R1 R2 R4 R7 R8 R7 = 7.7kQ R8 = 8.8k2 VI 20V R6 a. Number of nodes, branches, loops and mesh b. Equivalent resistance as seen at the voltage source c. Current through R1 and R4 d. Voltage across R5 and R8 e. Power at R6
- Finally, the multimeter is used to measure VOUT_C for the third potential divider. The circuit values are: VIN = 3.5 V, R1 = R2 = 6.6 kQ2, R3 = R4 = 660 kQ, and R5 = R6 = 6.6 MQ. VIN www www.ta R5 R6 VOUT C 1M Multimeter Your multimeter shows the measured value for VOUT_C to be 0.44 V. Consider the 3 resistors in the schematic above and recalculate what you expect the measured VOUT_C to be.How would I approach this question? Do I need to use Kirchoff's junction rules or can I just subtract the voltage drop?20 Resistors are said to be connected in series when two circuit elements connect at a single loop. Select one: True cion False
- Use the Superposition Principle to find the voltage VR5 (the voltage over the Resistor R5) - please only use the superposition principle rather than other methods as i am trying to learn how to use itValuesI1 = 5A V2 = 8V V3 = 9V V4 = 1VI5 = 2AI6 = 2AV7 = 10VR1 = 78ohms R2 = 54ohmsR3 = 5ohms R4 = 73ohms R5 = 88ohmsR6 = 42ohms R7 = 12 ohmsplease show complete solutionFor R1=285, R2=328, R3-414, R4=240, R5=160, R6-400, V1=3 V & I1=1 A in the shown circuit, using the superposition principle find the following: + V1 R5 Vo R4 + R6 w I1 R3 io R2 m ZR1 Iol (in ampere) due to V1 only= (Note that Io=Io1+Io2, where Iol is due to V1 and Io2 is due to I1) Vol (in volt) due to V1 only= (Note that Vo=Vol+Vo2, where Vol is due to V1 and Vo2 is due to I1) Io2 (in ampere) due to Il only= (Note that Io=Io1+Io2, where Iol is due to V1 and Io2 is due to I1) Vo2 (in volt) due to Il only= (Note that Vo=Vol+Vo2, where Vol is due to V1 and Vo2 is due to I1)