# 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 Capacitors Are Available to Him Each of Which Can - Physics

Numerical

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 capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

#### Solution

Total required capacitance, C = 2 µF

Potential difference, V = 1 kV = 1000 V

Capacitance of each capacitor, C1 = 1µF

Each capacitor can withstand a potential difference, V1 = 400 V

Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and the potential difference across each capacitor must be 400 V. Hence, the number of capacitors in each row is given as

1000/400 = 2.5

Hence, there are three capacitors in each row.

Capacitance of each row

= 1/(1 + 1 + 1)

= 1/3 mu "F"

Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

1/3 + 1/3 + 1/3 +..................  "n terms"

= "n"/3

However, the capacitance of the circuit is given as 2 mu"F"

∴ "n"/3 = 2

n = 6

Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.

Concept: Combination of Capacitors
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.23 | Page 89
NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 23 | Page 90

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