Calculate the equivalent capacitance of two capacitors C1 = 4 μF and C2 = 2 μF connected in series.

May 27, 2026Physics

ScanSolve AI Answer

May 27, 2026

Answer

4 , \mu\text{F} $and$ C_2 = 2 , \mu\text{F}$.

Here's the solution to the problem.

Step 1: Calculate the equivalent capacitance of the capacitors when connected in series. Given capacitances are $C_1 = 4 \, \mu\text{F}$ and $C_2 = 2 \, \mu\text{F}$ . For capacitors in series, the equivalent capacitance $C_{eq}$ is: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$ $\frac{1}{C_{eq}} = \frac{1}{4 \, \muF} + \frac{1}{2 \, \muF} = \frac{1+2}{4 \, \muF} = \frac{3}{4 \, \muF}$

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Here's the solution to the problem. Step 1: Calculate the equivalent capacitance of the capacitors when connected in series.

Asked on May 27, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI AnswerAnswered on May 27, 2026

Here's the solution to the problem. Step 1: Calculate the equivalent capacitance of the capacitors when connected in series. Given capacitances are C_1 = 4 \, and C_2 = 2 \, . For capacitors in series, the equivalent capacitance C_eq is: (1)/(C_eq) = (1)/(C_1) + (1)/(C_2) (1)/(C_eq) = (1)/(4 \, ) + (1)/(2 \, ) = (1+2)/(4 \, ) = (3)/(4 \, ) C_eq = \