Question

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

Short answer

The best arrangement is option D (6 rows with 3 capacitors), as it meets the required capacitance of 2 μF and stays within the 400 V limit for each capacitor. \(C_{total} = 2C\) and \(V_{cap} = \frac{V_{total}}{3}\).

Step-by-step solution

Option A - 2 rows with 2 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: \[V_{cap} = \frac{V_{total}}{2}\]

Option B - 4 rows with 2 capacitors

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: \[V_{cap} = V_{total}\]

Option C - 3 rows with 4 capacitors

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: \[V_{cap} = \frac{V_{total}}{4}\]

Option D - 6 rows with 3 capacitors

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: \[V_{cap} = \frac{V_{total}}{3}\]

Compare the arrangements

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.

Key concepts

Series and Parallel Circuits

Understanding the concepts of series and parallel circuits is crucial when working with capacitors. In a series circuit, capacitors are connected end-to-end, forming a single path for the electricity to flow. This setup results in a decrease in the total capacitance of the circuit. The formula to calculate the total capacitance in series is:

• The reciprocal of the total capacitance equals the sum of the reciprocals of each individual capacitance: \[ C_{total} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}} \]

In a parallel circuit, on the other hand, capacitors are connected in a manner that provides multiple paths for the current. This results in a higher total capacitance, as capacitors work together to store more charge. The calculation for capacitors in parallel is simpler:

• The total capacitance is the sum of each individual capacitance: \[ C_{total} = C_1 + C_2 + \ldots + C_n \]

These configurations are essential for designing circuits with specific requirements, such as achieving a desired capacitance or voltage handling capacity.

Capacitance Calculations

Capacitance calculations are the backbone of ensuring capacitors in a circuit meet the required specifications. When arranging capacitors, it is vital to achieve the desired total capacitance to ensure proper circuit functionality. In our scenario, the requirement is to obtain a total capacitance of \(2 \mu F\) using \(1 \mu F\) capacitors.
To achieve this in a series-parallel arrangement, two primary strategies help simplify calculations:

• For capacitors in series, the total capacitance is less than any individual capacitance because it creates a single pathway, distributing the charge across all units.
• For capacitors in parallel, their capacitances add up, enhancing the circuit's capacity to store charge and potentially meeting the required total capacitance.

In option D from the problem, we placed 3 capacitors in series in each row (equalizing half the requisite total capacitance), and then formed a parallel grouping with 6 such rows. This configuration smartly combines the advantages from both series and parallel arrangements to meet the \(2 \mu F\) requirement.

Voltage Distribution in Circuits

Handling voltage distribution correctly is critical when arranging capacitors in series and parallel configurations. Each capacitor can only withstand a certain voltage; thus, it's important to manage how voltages are shared among them. For capacitors in a series, the total voltage from the power source is divided equally among the capacitors.
Using the formula:

• Voltage across each capacitor = \( \frac{V_{total}}{n} \)

Where \( n \) is the number of capacitors in series.
For capacitors in parallel, each capacitor experiences the full voltage from the source, as they each connect directly to both sides of the voltage input:

• Voltage across each capacitor is equal to the total voltage: \[ V_{cap} = V_{total} \]

In option D, we effectively ensured each capacitor only needed to handle a third of the total voltage by placing them in rows of three in series. This technique allows us to safely operate within the capacitors' voltage ratings while achieving the desired overall circuit voltage.