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PROGRAM ROZWOJOWY

^1 POLITECHNIKI WARSZAWSKIEJ

To solve above eąuation one must find transient current and stationary current (Fig. 1.13). Stationary current satisfies eąuations:

^ = 0,andi= —.    (1.32)

dt    ^w R

Transient current satisfies eąuation:

Ri,+L^ = 0,    (1.33)

dt

which solution is as follow:

i_s = Ae ^L .

From boundary conditions: i(0) = i_w(0) + i_s(0) -> i_s(0) = -U/R = A Finally:

(1.35)

where: R/L - time constant (snubbing coefficient). Disconnecting process is shown in Fig.l. 14.

SW

(1.34)

i R    L

Fig. 1.14. Disconnecting process.

To avoid electric arc Circuit inductance must be short circuited. General law that governs this process is as follow:

Ri +    - o.    (1.36)

dt

and then:

(1.37)

UNIA EUROPEJSKA

EUROPEJSKI FUNDUSZ SPOŁECZNY

U. = ^U_R =L—5- = -Ue ’ dt

where: r = L/R.

a

Materiały dydaktyczne dystrybuowane bezpłatnie.

Projekt współfinansowany ze środków Unii Europejskiej w ramach Europejskiego Funduszu Społecznego

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