Question

A=63.0 B=73.0 C=173.0 D=10.0 anected across capacitors of capacitances ( C₁=C₁=C_4=C₃ =D mu F₂ C₂=fracD2 mu F ). Calculate the equivalent capacitance Cen of the capacitors.

Answer 1

Firstly, the capacitances of the capacitors are given as follows:

C₁ = D = 10.0 µF
C₂ = D/2 = 5.0 µF
C₃ = D = 10.0 µF
C₄ = D = 10.0 µF

To find the equivalent capacitance of series-connected capacitors, we need to use the formula for capacitors in series. Using this formula, the equivalent capacitance is given by the reciprocal of the sum of the reciprocals of individual capacitances. Formally, this can be written as:

1/C_eq_series = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄

Substituting the values of C₁, C₂, C₃, and C₄ into this formula, we get:

1/C_eq_series = 1/10.0 + 1/5.0 + 1/10.0 + 1/10.0 = 0.1 + 0.2 + 0.1 + 0.1 = 0.5 (Note that the units µF in the fractions cancel out)

However, 0.5 is not the equivalent capacitance of the capacitors but the reciprocal of it. To find the equivalent capacitance (C_eq), we need to take the reciprocal of 1/C_eq_series:

C_eq = 1 / C_eq_series = 1 / 0.5 = 2.0 µF

Therefore, the equivalent capacitance of the series-connected capacitors is 2.0 microfarads (µF).