An electrical technician requires a capacitance of \(2 \mu \mathrm{F}\) in a circuit across a potential difference of \(1 \mathrm{KV}\). A large number of $1 \mu \mathrm{F}$ capacitors are available to him, each of which can withstand a potential difference of not than \(400 \mathrm{~V}\). suggest a possible arrangement that requires a minimum number of capacitors. (A) 2 rows with 2 capacitors (B) 4 rows with 2 capacitors (C) 3 rows with 4 capacitors (D) 6 rows with 3 capacitors

In this arrangement, we have 2 rows of capacitors connected in series, with each row containing 2 capacitors. The total capacitance for capacitors connected in series is given by: \[C_{total} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}}\] Since both capacitors have the same capacitance, the formula becomes: \[C_{total} = \frac{1}{2C}\] Then, we need to check how much voltage each capacitor can handle. For series connection, the voltage is divided equally among the capacitors, so each capacitor will handle half of the potential difference: \[V_{cap} = \frac{V_{total}}{2}\]

In this arrangement, the total capacitance for each row of capacitors connected in series would be the same as in step 1: \[C_{total} = \frac{1}{2C}\] However, since we have 4 rows of capacitors connected in parallel, the total capacitance becomes: \[C_{total} = 4 \times \frac{1}{2C} = 2C\] The voltage for capacitors in parallel connections remains the same, so each capacitor will handle the full potential difference: \[V_{cap} = V_{total}\]

In this arrangement, we have 3 rows of capacitors connected in parallel, with each row containing 4 capacitors in series. For the total capacitance in each row, we have: \[C_{row} = \frac{1}{4C}\] Since we have 3 rows of capacitors connected in parallel, the total capacitance becomes: \[C_{total} = 3 \times \frac{1}{4C} = \frac{3}{4C}\] The voltage for capacitors in series is divided equally among the capacitors, so each capacitor will handle a quarter of the potential difference: \[V_{cap} = \frac{V_{total}}{4}\]

In this arrangement, we have 6 rows of capacitors connected in parallel, with each row containing 3 capacitors in series. For the total capacitance in each row, we have: \[C_{row} = \frac{1}{3C}\] Since we have 6 rows of capacitors connected in parallel, the total capacitance becomes: \[C_{total} = 6 \times \frac{1}{3C} = 2C\] The voltage for capacitors in series is divided equally among the capacitors, so each capacitor will handle one-third of the potential difference: \[V_{cap} = \frac{V_{total}}{3}\]

Now we will compare the total capacitance and voltage handled by each capacitor for the given arrangements. Option A: - Total capacitance: \(\frac{1}{2C}\) (falls short of required capacitance) - Voltage per capacitor: \(V_{total}/2\) (exceeds 400 V limit) Option B: - Total capacitance: \(2C\) (meets required capacitance) - Voltage per capacitor: \(V_{total}\) (exceeds 400 V limit) Option C: - Total capacitance: \(\frac{3}{4C}\) (falls short of required capacitance) - Voltage per capacitor: \(V_{total}/4\) (within 400 V limit) Option D: - Total capacitance: \(2C\) (meets required capacitance) - Voltage per capacitor: \(V_{total}/3\) (within 400 V limit) Based on the comparison, option D (6 rows with 3 capacitors) is the best arrangement, as it meets both the required capacitance and voltage limits.