JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 12)
Two identical capacitors A and B, charged to
the same potential 5V are connected in two
different circuits as shown below at time t = 0.
If the charge on capacitors A and B at time
t = CR is QA and QB respectively, then (Here
e is the base of natural logarithm)
_9th_January_Evening_Slot_en_12_1.png)
QA = $${{CV} \over e}$$, QB = $${{VC} \over 2}$$
QA = $${{CV} \over 2}$$, QB = $${{VC} \over e}$$
QA = VC, QB = $${{VC} \over e}$$
QA = VC, QB = CV
Explanation
In case I diode is reverse biased
QA = CV
In case II, current will flow as diode is forward biased.
QA = $${{CV} \over e}$$
QA = CV
In case II, current will flow as diode is forward biased.
QA = $${{CV} \over e}$$
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