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What is the maximum number of 7.00 μF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC ?

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Nearly 45 capacitors are required.

Answer:

Step-by-step explanation:

It is known that voltage flow in any parallel circuit is tend to be constant. And here the circuit is made up of paralleled arranged capacitors. A capacitor is the one which generates current for a given voltage in the circuit. So the mathematical representation of capacitance is

C = I/V, here I is the current and V is the voltage.

So if capacitors are connected in parallel, then the sum of varying current will give the equivalent current in the circuit.

I = I1+I2+....In

So,
C_(p) V_(p)=C_(1) V_(1)+ C_(2) V_(2)+ C_(3) V_(3) +......+C_(n) V_(n)

Since, in the present case, the capacitance is kept constant for all the capacitors connected in parallel and the voltage is constant in parallel circuit connection, then

I= nCV

n = I/CV=953/(3×7) = 45

So nearly 45 capacitors are required.

User Javier Ramirez
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