Q58PE - College Physics (Urone)

Question

Suppose you want a capacitor bank with a total capacitance of 0.750F and you possess numerous 1.50mF capacitors. What is the smallest number you could hook together to achieve your goal, and how would you connect them?

Step-by-Step Solution

Answer

 500 capacitors need to hooked in parallel to form a capacitor bank with a total capacitance of 0.750 F.

Step 1: Concept Introduction

Capacitors in Parallel: When capacitors with capacitances  C₁, C₂, …. are connected in parallel, the equivalent capacitance C_eq equals the sum of the individual capacitances –

 

Step 2: Calculation of Capacitance

Since equivalent capacitance of the bank is more than capacitance of each capacitor, the capacitors must be connected in parallel. The equivalent capacitance of n capacitors that are connected in parallel is-

 ${\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3+\dots +{\mathrm{C}}_{\mathrm{n}}$

Here ${\mathrm{C}}_1={\mathrm{C}}_2={\mathrm{C}}_3={\mathrm{C}}_{\mathrm{n}}=\mathrm{C}$ because the capacitors are identical, therefore, we can write the above sum in terms of the number n of the connected capacitors:

 ${\mathrm{C}}_{\mathrm{eq}}=\mathrm{nC}$

We need ${\mathrm{C}}_{\mathrm{eq}}$ to be 0.750 F and C = 1.50 mF is the capacitance of the small capacitors. Solving for n, it is obtained –

 $\begin{array}{rcl}\mathbf{n}& \mathbf{=}& \frac{{\mathbf{C}}_{\mathbf{eq}}}{\mathbf{C}}\\ & \mathbf{=}& \frac{\mathbf{0}\mathbf{.}\mathbf{750}\mathbf{ }\mathbf{F}}{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{mF}\mathbf{)}\left(\frac{1 F}{1000\mathrm{mF}}\right)}\\ & \mathbf{=}& \mathbf{500}\end{array}$

 

Therefore, you need to connect 500 capacitors in parallel to form a capacitor bank with a total capacitance of 0.750 F .