JAMB Physics 2024 Paper

Combined capacitance in series = $\frac{1}{C_T}$ = $\frac{1}{C_1}$ + $\frac{1}{C_2}$

The two capacitors have capacitance of 0.0003μF and 0.0006μF respectively

$\frac{1}{C_T}$ = $\frac{1}{0.0003}$ + $\frac{1}{0.0006}$ = $\frac{0.0002 + 0.0001}{0.0006}$

$\frac{1}{C_T}$ = $\frac{0.0003}{0.0006}$

C${}_T$ = 0.0002μF.