If you have several `2.0 muF` capacitrors, each capable of withstanding `200 V` without breakdown, how would you assemble a combination that has an equivalent capacitance of
a. `0.4muF`
b. `1.2 muF` each with standing `1000V`?

Several 10 pF capacitors are given, each capable of withstanding 100 V . How would you construct : (a) a unit possessing a capacitance of 2 pF and capable of withstanding 500 V ? (b) a unit possessing a capacitance of 20 pF and capable of withstanding 300 V ?

Several 10 pF capacitors are given, each capable of withstanding 100 V . How would you construct : (a) a unit possessing a capacitance of 2 pF and capable of withstanding 500 V ? (b) a unit possessing a capacitance of 20 pF and capable of withstanding 300 V ?

You are provided with 3 capacitors, each capacitance 2muF . In how many possible ways can you connect these capacitors. Calculate the capacitance of each of the combinations.

The minimum number of capcitors each of 3muF required to make a circuit with an equivalent capacitance 2.25muF is

Three capacitors of capacitances [2muF, 3muF and 4muF] are connected in sereis. : a Find the equivalent capacitance of the combination

Four capacitors of value 6muF, 6muF, 6muF and 2muF are connected to a 6 V battery as shown in the fig Determine the equivalent capacitance of the network

Find the equivalent capacitance between points A and B. Capacitance of each capacitor is 2muF .

Find the equivalent capacitance between points A and B. Capacitance of each capacitor is 2muF .

An infinite number of capacitors, having capacitances 1muF, 2muF, 4muF and 8muF….. are connected in series. The equivalent capacitance of the system is