CBSE Class 12-science Answered
In the circuit, shown in figure 'K' is open. The charge on capacitor C in steady state is q1. Now key is closed and at steady state, the charge on C is q2. The ratio of charges is
Asked by Atulcaald | 19 May, 2018, 12:59: AM
Expert Answer
When the key is open, after applying the EMF from battery through resistance, capacitor is getting charged.
When it reaches steady state, charge q1 on capacitor is given by : q1 = C×E ..............(1)
C is capacitance in Farad and E is EMF of battery in Volts.
After reaching steady state, when key is closed, current will not pass through capacitors from battery because it is fully charged.
We can assume the capacitor is open.
Now potential difference across the plates of capcitor is same as the potential difference across the resistor that has the value of 2R Ω.
The potential difference across 2R Ω is (2/3)E. Hence charge q2 across the plates of capacitor becomes
q2 = C×(2/3)E ................................(2)
from (1) and (2), q1/q2 = 3/2
Answered by Thiyagarajan K | 20 May, 2018, 01:02: PM
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