As an electrical engineer, you're asked to specify a capacitor that can store 12 mJ of energy. The largest capacitor that will physically fit on your circuit board is \(10 \mu \mathrm{F}\). The manufacturer produces capacitors with voltage ratings in multiples of \(25 \mathrm{V}\). What voltage do you specify?

Short Answer

Expert verified
The voltage to be specified is 500V. This is the closest increment of 25V to the calculated voltage value.

Step by step solution

01

Convert the energy into Joules

Before getting started, we have to make sure that all the quantities involved are in the same unit. In this case, energy is given in milli Joules (mJ) and needs to be converted into Joules (J) since we use SI units in our calculations. 1 Joule = 1000 milli Joules. So, 12mJ = 12/1000 J = 0.012 J.
02

Find voltage using the formula

Now that we have the energy in the correct units, we can use the energy formula of a capacitor which is \(U = \frac{1}{2}CV^2\). Here, U is the stored energy, C is the capacitance and V is the potential difference or voltage. We have to solve for V. So, rearrange the formula to get \(V = \sqrt{\frac{2U}{C}}\). Plug in the given values, \(V = \sqrt{\frac{2*0.012 J}{10*10^-6 F}}\).
03

Find the next highest 25V increment

After calculating, we get a voltage slightly less than 500V. However, the manufacturer produces capacitors with voltage ratings in multiples of 25V so we need to choose the next highest multiple of 25. Hence, we round up to 500V as our final answer.

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