A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1 . 0 kV . A large number of 1 μ F capacitors are available which can withstand a potential difference of not more than 3 0 0 V . The minimum number of capacitors required to achieve this is:

Step-by-step solution

Let us assume that we connect n (whole number) capacitors in series such that the potential difference across the combination is 1000   V . This means the potential difference across each capacitor is V c a p a c i t o r = 1000 n   V ≤ 300   V n ⩾ 1000 300 = 3 . 333 . n m i n = 4 This means we need to connect at least 4 capacitors in series to make sure that the potential across each one is less than 300   V . C e q = m n C = 2   μF m 4 = 2 ⇒ m = 8 ∴ Minimum no