An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitorsare available to him each of which can withstand a potentialdifference of not more than 400 V. Suggest a possible arrangementthat requires the minimum number of capacitors.
Given: The required capacitance is 2 μF , the potential difference is 1 kV and the capacitance of each capacitor is 1 μF the capacity of potential difference is 400 V .
The number of capacitor in each row is given as,
n = V V 1
Where, the potential difference across the circuit is V and the capacity of potential difference on a capacitor is V 1 .
By substituting the given values in the above formula, we get
n = 1 kV 400 V = 1000 V 400 V = 2.5
So, the approx value of number of capacitor is 3.
The capacitance of each row is given as,
1 C = 1 C 1 + 1 C 2 + 1 C 3
Where, the capacitance of the capacitors are C 1 , C 2 and C 3 .
By substituting the given values in the above formula, we get
C = 1 1 + 1 + 1 = 1 3 μF
The equivalent capacitance of the circuit is given as,
C n = n 3
Where, the numbers of rows are n .
By substituting the given values in the above formula, we get
2 = n 3 n = 6
Thus, in the circuit the rows of three capacitors are 6 and the for the given arrangement the required capacitors are 18 .
Given: The required capacitance is
The number of capacitor in each row is given as,
Where, the potential difference across the circuit is
By substituting the given values in the above formula, we get
So, the approx value of number of capacitor is 3.
The capacitance of each row is given as,
Where, the capacitance of the capacitors are
By substituting the given values in the above formula, we get
The equivalent capacitance of the circuit is given as,
Where, the numbers of rows are
By substituting the given values in the above formula, we get
Thus, in the circuit the rows of three capacitors are
