Q. 144.8( 5 Votes )

It is required to construct a 10 μF capacitor which can be connected across a 200V battery. Capacitors of capacitance 10 μF are available, but they can withstand the only 50V. Design a combination which can yield the desired result.

Answer :

Requirement : We have to construct a 10μF capacitor, and it has to connect across a 200V battery.


Capacitors of 10μF are available, but the voltage rating is 50V only. By using these capacitors with this voltage rating, we have to meet our requirement.


Let's assume some X capacitors are placed in series. 200V battery connected across the.


So




We have to construct 4 capacitors in a series so that we get the potential difference of 200V


If we calculate the capacitance of the parallel combination of four 10μF capacitors



Putting the value of the capacitor in the above formula, we get



Thus, the capacitance of the combination is C=2.5μF


With this arrangement, we get the required potential difference value, but we are not getting the capacitor value 10μF instead of this we get only 2.5μF. So we have to add some columns. Let us take Y as columns,




So we have to add 4 columns as the same row.


That circuit will look like



Finally, the above fig will be the design for our requirements; each capacitor value is with voltage rating 50V.


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In the circuit shown in Fig. 2.4. initially key K1 is closed and key K2 is open. Then K1 is opened and K2 is closed (order is important).

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