134)seems dat a bit tough one..........
i think we ve to apply many principle in it ....(considerin d inductor alone we ve to make use of inductive reactance n also there must be an in duced emf too)
wait pls.........
10 Answers
11post d solution without fail after u get it...........
byah.. need ur help.........
130.
The positive or negative half-cycle
of Vin (t) [which is, here, V(t)] determines
whether the diodes are ON
or OFF (respectively).
The maximum values of Vo(t) are
determined purely by the voltage
remaining after the diode drop of
the arm powered by the dc
source.
During positive half-cycle of Vin(t),
we have D1 as ON-diode.
⇒ Vo (max) = V1 − Vdiode-drops
= 2 − 0.7
= 1.3 V
When D2 is ON (i.e., during
negative half of V(t))
We have Vo (min) = − (1 − 0.7)
= − 0.3 V
1wat is the q in 30th one.....
as the diode is reverse biased., no current flows through the diode.... now it becomes an AC current quesion.. but if the switch is on n off at an interval, then there will be charging and discharging....etc i think i will b complicated