How do I solve for the values calculated from the postulated fault, which has to

Question

How do I solve for the values calculated from the postulated fault, which has to be determined by the differences between the circuit without faults and the circuit with a fault?

experiment, only a single type of fault will be induced to a single el circuit. ement of the In many cases, the faults are not o order to find the faults, one must "troubleshoot" the circuit. That is, one resistance, voltage and/or current at selected points in the circuit an measurements, postulates possible faults. The methods of D.C. circuit analysis, as bvious when the circuit operates improperly. In measures the d from these described above can be used to verify or disprove these postulates. o@ R1:400 R2 = 200 E S1 B S2 Roar-R3 each resistor V2 -| is 100 Figure 18 -5 Figure 18-5 shows the circuit you will be dealing with in this lab. Rpar is made up of three 100 resistors in parallel. RF represents the resistance imposed upon the circuit by a resettable fuse and also the meter when measuring current.

Explanation / Answer

Since RF is imposed on the circuit, difference in the values of resistances and currents occurs due to this resistor and hence it represents a fault in the given circuit. The measured values from postulated errors are:

So, R1 =400+ 400*8/(400+8) = 407.84 ohm

Similar;y, R2  = 200 + 200*8/(200+8) = 207.69 ohm

R3  = 33.33 +33.33*8/(33.33+8) = 39.78 ohm

RBC = 400+33.33+ 433.33*8/(433.33+8) =441.185 ohm

RBD = 400+200+ 600*8/(600+8) = 607.89 ohm

RDC = 200+33.33+ 233.33*8/(233.33+8) = 241.065 ohm

Since the calculated volatge is 3 Volts, So

I1 = V/R1 = 3/407.84 = 7.36 mA

I2 = V/R2 = 3/207.69 = 14.44 mA

I3 = V/(R3+RF) = 3/(39.78+8) = 0.0628 A

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