The LR circuit below contains a small resistance r =1 Ω and an inductance L = 1 mH in series with a baiery of
emf ε0= 2V. The switch S is iniPally closed. At t=0 the switch is opened so that the large circuit breaker resistance
Rcb= 999 Ω is now in series with the other circuit elements.
(a) If S had been in the closed posiPon for a long Pme before t=0 , what was the steady current I0 flowing in
the circuit for t < 0 ?
(b) With the steady current I0 flowing, the switch S is opened at t=0 so that the circuit resistance is now
R =Rcb+r = 1kΩ. Write the differenPal equaPon for I(t) that describes the change in the current for t ≥ 0 .
Solve this equaPon (by integraPon) for I(t) under the approximaPon that ε0=0. (The baiery emf will be
negligible compared to the potenPal difference across the inductor for Pmes just afer the switch is closed, as
can be verified from the answer.) Use the values provided, including the result found in part (a), to replace R,
L and I0 by numerical values when wriPng down your final expression for I(t) In the answer sheet.
(c) Hence find the potenPal drop across the circuit resistance of R = 1 kΩ just afer the switch is opened. You
should find that it greatly exceeds ε0= 2V.
r =1Ω
Rcb=999 Ω
ε0= 2V
L=10−3 H
please help me!
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