1) A circuit is made with 128 V battery and four identical capacitors (C = 2.00mF) as shown Find the charge and voltage on each capacitor i) With Switch S OPEN ii) With Switch S CLOSED C C
Equivalent capacitance of capacitors connected in parallel is given by
Equivalent capacitance of capacitors connected in series is given by
Wheatstone bridge principle states that if the ratio of , then the bridge is said to be balanced and the potential at points A and B is same and thus no current flows in the arm AB.
(a)
When switch S is open-
Equivalent capacitance of upper two capacitors in series are-
Similarly the net capacitance of lower two capacitors is also
Therefore, the equivalent capacitance of the circuit is given by-
So total charge flowing in the circuit is-
Since the net capacitance in upper arm and lower arm is same, so the charge gets divided equally between both the arms.
So, total charge flowing in the combination of lower arm of capacitors is-
This is also the charge present on each of the lower two capacitors.
Similarly the charge on each of the upper two capacitors is also 128 mC
Now, voltage present on each of the lower capacitor is given by-
Similarly the voltage across each of the capacitors present in upper arm is also 64 Volts
(b)
When switch S is closed, it becomes a balanced condition of Wheatstone bridge and thus the potential difference between the two ends of the middle arm is zero.
It is now the similar case as of part (a). Hence charge on each of the capacitor is 128 mC and voltage on each of the capacitor is 64 Volts.
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