A step voltage `V_(0)` is applied to a series combination of `R` and `C` as shown Fig. Then A. after sufficienty long time `V_(R ) = 0`B. as time passes, `V_(R )` decreases as `(1//t)`C. after `1 ms, V_(C ) = 6.3` volt (approximately)D. initially, current through `R` is `10 mA`.

A step voltage `V_(0)` is applied to a series combination of `R` and `

1.	A step voltage `V_(0)` is applied to a series combination of `R` and `C` as shown Fig. Then A. after sufficienty long time `V_(R ) = 0`B. as time passes, `V_(R )` decreases as `(1//t)`C. after `1 ms, V_(C ) = 6.3` volt (approximately)D. initially, current through `R` is `10 mA`.
Answer» Correct Answer - A::C::D (a) After sufficently long time, since the charging current drops to zero, voltage drop across resistance is zero. (c ) Time constant ` = RC = K Omega xx 1 mu F = 1 ms`. So, accoridng to the definition of time constant, voltage across capacitor would be about `63%` of maximum voltage, i.e., `6.3 V` after 1 ms. (d) Initital current `= (10 V )/(1 KOmega) = 10 mA`

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A step voltage `V_(0)` is applied to a series combination of `R` and `C` as shown Fig. Then A. after sufficienty long time `V_(R ) = 0`B. as time passes, `V_(R )` decreases as `(1//t)`C. after `1 ms, V_(C ) = 6.3` volt (approximately)D. initially, current through `R` is `10 mA`.

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