Four 16 micro F capacitors are connected in combination. What is the equivalent capacitance of this combination if they are connected (a) in series, (b) in parallel, and (c) such that two of them are

(a) When capacitors are connected in series, their equivalent capacitance can be calculated using the formula: {eq}frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + dotsb + \frac{1}{C_n} {/eq} where {eq}C_{eq}{/eq} is the equivalent capacitance, and {eq}C_1, C_2, dotsb, C_n{/eq} are the capacitances of the individual capacitors. In this case, since there are four capacitors of equal capacitance, we can simplify the formula as: {eq}frac{1}{C_{eq}} = \frac{1}{16 mu F} + \frac{1}{16 mu F} + \frac{1}{16 mu F} + \frac{1}{16 mu F} {/eq} which gives: {eq}frac{1}{C_{eq}} = \frac{4}{16 mu F} {/eq} Simplifying further, we get: {eq}C_{eq} = \frac{16 mu F}{4} {/eq} Therefore, the equivalent capacitance of the combination when the capacitors are connected in series is: {eq}C_{eq} = 4 mu F {/eq} (b) When capacitors are connected in parallel, their equivalent capacitance can be calculated simply by adding up the capacitances of the individual capacitors. In this case, since all four capacitors are of equal capacitance, their equivalent capacitance is: {eq}C_{eq} = 16 mu F + 16 mu F + 16 mu F + 16 mu F {/eq} which gives: {eq}C_{eq} = 64 mu F {/eq} Therefore, the equivalent capacitance of the combination when the capacitors are connected in parallel is: {eq}C_{eq} = 64 mu F {/eq} (c) When two capacitors are connected in parallel, their equivalent capacitance is the sum of their individual capacitances. We can simplify the circuit as follows: {eq}begin{aligned} & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \frac{1}{16 mu F} \quad \quad \frac{1}{16 mu F} \quad \quad \quad \quad \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \frac{1}{16 mu F} \quad \quad \frac{1}{16 mu F} \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \frac{1}{16 mu F} \quad \quad \quad \quad \quad \quad \quad \quad \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \frac{1}{16 mu F} \quad \quad \frac{1}{16 mu F} \quad \quad \quad \quad \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad & \ end{aligned} {/eq} The two capacitors in parallel have an equivalent capacitance of: {eq} C_{eq1} = 16 mu F + 16 mu F = 32 mu F {/eq} The remaining two capacitors are in series, which we have already calculated in part (a) to be: {eq} C_{eq2} = 4 mu F {/eq} The equivalent capacitance of the entire combination is the sum of {eq}C_{eq1}{/eq} and {eq}C_{eq2}{/eq}, which gives: {eq}C_{eq} = 32 mu F + 4 mu F {/eq} Therefore, the equivalent capacitance of the combination when two capacitors are in parallel and that combination is in series with the remaining two capacitors is: {eq}C_{eq} = 36 mu F {/eq}