2.23 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.

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alok kumar singh | Contributor-Level 10

3 months ago

2.23 Requirements:

Capacitance, C = 2 $μ F$

Potential difference, V = 1 kV = 1000 V

Available:

Each capacitor, capacitance, $C_{1}$ = 1 $μ F$

Potential difference, $V_{1}$ = 400 V

Assumption:

A number of capacitors are connected in series and these series circuits are connected in parallel to each other. The potential difference across each row in parallel connection must be 1000 V and potential difference across each capacitor must be 400 V.

Hence, number of capacitors in each row is given as $\frac{1000}{250}$ = 2.5

So the number of capacitor in each row = 3

The equivalent capacitance in each row is given as, $\frac{1}{C_{R}}$ = $\frac{1}{1} + \frac{1}{1}$ + $\frac{1}{1}$

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2.23 Requirements:

Capacitance, C = 2 $μ F$

Potential difference, V = 1 kV = 1000 V

Available:

Each capacitor, capacitance, $C_{1}$ = 1 $μ F$

Potential difference, $V_{1}$ = 400 V

Assumption:

Hence, number of capacitors in each row is given as $\frac{1000}{250}$ = 2.5

So the number of capacitor in each row = 3

The equivalent capacitance in each row is given as, $\frac{1}{C_{R}}$ = $\frac{1}{1} + \frac{1}{1}$ + $\frac{1}{1}$ = 3, so $C_{R} = \frac{1}{3}$ $μ$ F

Let there be ‘n’ rows, each having 3 capacitors, which are connected in parallel

The total equivalent capacitance of the circuit is given as

$C_{e f f}$ = $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + ……. upto n = $\frac{n}{3}$

It is given $C_{e f f}$ = 2. So n = 6

So the final circuit will be 3 capacitors in each row and 6 rows connected in parallel, Total capacitors required = 3 $\times 6 = 18$

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2.23 Requirements:Capacitance, C = 2 <math><mi>μ</mi><mi>F</mi></math>Potential difference, V = 1 kV = 1000 VAvailable:Each capacitor, capacitance, <math><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 1 <math><mi>μ</mi><mi>F</mi></math>Potential difference, <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 400 VAssumption:A number of capacitors are connected in series and these series circuits are connected in parallel to each other. The potential difference across each row in parallel connection must be 1000 V and potential difference across each capacitor must be 400 V.Hence, number of capacitors in each row is given as <math><mfrac><mrow><mrow><mn>1000</mn></mrow></mrow><mrow><mrow><mn>250</mn></mrow></mrow></mfrac></math> = 2.5So the number of capacitor in each row = 3The equivalent capacitance in each row is given as, <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>R</mi></mrow></mrow></msub></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></mfrac><mo>+</mo><mi></mi><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></mfrac></math> + <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></mfrac></math> = 3, so <math><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>R</mi></mrow></mrow></msub><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> <math><mi>μ</mi></math> FLet there be ‘n’ rows, each having 3 capacitors, which are connected in parallelThe total equivalent capacitance of the circuit is given as<math><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>e</mi><mi>f</mi><mi>f</mi></mrow></mrow></msub></math> = <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> + <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> + <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> + <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math> + ……. upto n = <math><mfrac><mrow><mrow><mi>n</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math>It is given <math><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>e</mi><mi>f</mi><mi>f</mi></mrow></mrow></msub></math> = 2. So n = 6So the final circuit will be 3 capacitors in each row and 6 rows connected in parallel, Total capacitors required = 3 <math><mo>×</mo><mn>6</mn><mo>=</mo><mn>18</mn></math>

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