A capacitor 1 with breakdown voltage 120 volt is connected with two other capacitors 2 and 3 (2 and 3 are in parallel connected with 1 as series) .the breakdown voltage of 2 is 120 V but that of 3 is 150 V. The maximum potential across the plates of 2 so that all capacitors remain safe (all capacitors have same capacity)
Answers
Maximum potential difference that can be obtained across the plates of capacitor 2 is 60 Volts , in order to keep all the three capacitors safe.
Explanation:
Given: A capacitor 1 with breakdown voltage 120 volt is connected with two other capacitors 2 and 3 (2 and 3 are in parallel connected with 1 as series). The breakdown voltage of 2 is 120 V but that of 3 is 150 V. All capacitors have same capacity.
Find: The maximum potential across the plates of 2 so that all capacitors remain safe.
Solution:
Let us assume that breakdown voltage of approx. 120 V has been obtained on the plates of capacitor 1.
Charge on capacitor 1 = q = CV = C * 120
Capacitors 2 and 3 are in parallel connected with 1 as series.
Equivalent capacitance (considering capacitor 2 and 3) = (C + C) = 2C.
So potential difference across 2C = V = q/C = 120 / 2 = 60 volt.
Maximum potential difference that can be obtained across the plates of capacitor 2 is 60 Volts , in order to keep all the three capacitors safe.