Question

what is the capacitance in a circuit that has three capacitors set up in series? c1 = 0.50 f, c2 = 0.50 f, and c3 = 0.50 f. o 0.17 f o 6.0 f o 1.5 f o 0.13 f save for later 27.4 rc circuits submit ans

Explanation:

Step1: Recall series - capacitance formula

For capacitors in series, the formula is $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$.

Step2: Substitute given values

Given $C_1 = 0.50F$, $C_2 = 0.50F$, $C_3 = 0.50F$. Then $\frac{1}{C_{eq}}=\frac{1}{0.50}+\frac{1}{0.50}+\frac{1}{0.50}$.

Step3: Calculate the right - hand side

$\frac{1}{0.50}+\frac{1}{0.50}+\frac{1}{0.50}=2 + 2+2=6$. So, $\frac{1}{C_{eq}} = 6$.

Step4: Solve for $C_{eq}$

$C_{eq}=\frac{1}{6}\approx0.17F$.

Answer:

0.17 F