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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance5 Marks
Question
An electrical technician requires a capacitance of 2μF in a circuitacross a potential difference of 1kV. A large number of 1μF capacitors are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors.
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Answer

Total required capacitance, C = 2μF Potential difference, V = 1kV = 1000V Capacitance of each capacitor, C1 = 1μF

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

$\frac{1000}{400}={2.5}$

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 1000V and potential difference across each capacitor must be 400V. Hence, number of capacitors in each row is given as,

Hence, there are three capacitors in each row.

Capacitance of each row $=\frac{1}{1+1+1}=\frac{1}{3}\ \mu \text{F}\ \frac{1}{3}+\frac{1}{3}+\frac{1}{3}+ .....\text{n terms,}$ 

However, capacitance of the ciruit is given as 2μF.

$\therefore\ \frac{\text{n}}{3}={2}$

N = 6

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

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

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