Question

A fixed capacitor of capacitance 10^(-4) microfarad is connected in series with a variable capacitor, the capacitance of which may be varied from zero to 10^(-4) microfarad in steps of 10^(-6) microfarad. These two in series are connected in parallel with a third capacitor of capacitance 5 x10^(-4) microfarad. Calculate:
(i) the maximum capacitance of the whole combination.
(ii) the smallest change in this capacitance which can be produced by the arrangement.

Answer 1

Let's write the equivalent capacitance of the whole combination of capacitors:

$$C_{eq}=\dfrac{C_1C_2}{(C_1+C_2)}+C_3.$$

(i) As we can see from the formula, the maximum capacitance of the whole combination will be when $C_2=10^{-4}\ \mu F$:

$$C_{eq}=\dfrac{10^{-4}\ \mu F\cdot10^{-4}\ \mu F}{(10^{-4}\ \mu F+10^{-4}\ \mu F)}+5\cdot10^{-4}\ \mu F=5.5\ \mu F.$$

(ii) Let's substitute into $C_{eq}$ the capacitance of $C_2=10^{-6}\ \mu F$ (the smallest possible value):

$$C_{eq,1}=\dfrac{10^{-4}\ \mu F\cdot10^{-6}\ \mu F}{(10^{-4}\ \mu F+10^{-6}\ \mu F)}+5\cdot10^{-4}\ \mu F=5.0099\ \mu F.$$

Then, let's change it by step ($10^{-6}\ \mu F$) and again substitute into the formula:

$$C_{eq,2}=\dfrac{10^{-4}\ \mu F\cdot2\cdot10^{-6}\ \mu F}{(10^{-4}\ \mu F+2\cdot10^{-6}\ \mu F)}+5\cdot10^{-4}\ \mu F=5.0196\ \mu F.$$

Then, the smallest change in this capacitance which can be produced by the arrangement equals the difference between $C_{eq,2}$ and $C_{eq,1}$:

$$\Delta C_{eq}=C_{eq,2}-C_{eq,1},$$ $$\Delta C_{eq}=5.0196\ \mu F-5.0099\ \mu F=9.7\cdot10^{-3}\ \mu F.$$