The circuit of figure -3 below is intended for reduced...

The circuit of figure -3 below is intended for reduced voltage starting of a 3-phase induction motor using the wye-delta starting method. 4 1 2 STOP START PB1 P82 M1 HO HI X1 M2 TRI C1 C1 M2 OLS TRY (2-3) ON-DELAY TIMER TWI STA T10 T20 13 T a) From the control circuit, immediately after the START push button is pressed, which contact(s) closes or opens, and which coils would be energized or de-energized (fill in the info below). Closed contacts: Open contacts: Energized coils: De-energized coils: b) When would the C1 coil be de-energized during the starting of the motor, and why de-energize the coil? (Assume a timer setting of 10 seconds).

Transcript

00:01 In the question we are given this lc circuit, the values of l, c and q are given as l is equals to 3 .3 henry.

00:16 C is equals to 834 picofarad that is equals to 834 into 10 to the power minus 12 ferret.

00:26 Then charge q is given to be 127 microculems that is 127 into 10 to the power minus 6 coolums and time t is given to be 3 milliseconds that is equals to 3 into 10 to the power minus 3 seconds energy stored in the capacitor is given by half q square c now, omega is equals to 1 upon square root of lc.

01:07 So let us first find out omega.

01:10 Substituting the values we get 1 upon square root of 3 .3 into 834 into 10 to the power minus 12.

01:21 So on simplification we get 10 to the power 6 upon 52 .46.

01:30 Equals to 0 .019 into 10 to the power 6 radiance per second.

01:38 Now charge in lc circuit is given by q equals to q max cause omega t.

01:49 In the question we are given q max that is 127 microculems.

01:55 So on substituting the values we get 127 into 10 to the power minus 6 cause of 0 .019 into 10 to the power 6 into 3 into 10 to the power minus 3.

02:11 So this is equals to 127 into 10 to the power minus 6 cause of 57 degree.

02:20 That is equals to 127 into 0 .54 into 10 to the power minus 6 that is equals to 68 .4 .4 into 10 to the power minus 6.

02:31 58 microculems so this is the charge q now let us substitute the value of q and see in equation 1 so we get ec is equals to half of 68 .58 into 10 to the power 6 the whole square upon 834 into 10 to the power minus 12 so on simplification we get ec is 2 .82 joules.

03:07 Thus, the energy stored in the capacitor is 2 .82 joules...

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