The equation for charge $Qon$ a capacitor while charging $is$ given $by$:

$Q=C{V}_b(1-e^{-\frac{t}{R_{C}C}})$

where $V_{b}is$ the voltage $of$ the battery, $R_{c}is$ the total resistance $in$ the

charging circuit, $tis$ the time since the capacitor started $to$ charge and $Cis$

the capacitance.

The equation for the charge $on$ the capacitor while discharging $is$:

$Q=Q_0{e}^{-\frac{t}{R_{d}C}}$

where $Q_{0}is$ the charge $on$ the capacitor just before $it$ starts $to$ discharge, $R_d$

$is$ the total resistance $in$ the discharging circuit and $t$ now represents the

time since the capacitor started $to$ discharge.

With $R1=5andR2=5,$ the charge $on$ the capacitor $att=10sis...(to3dp)$

The equation for charge Q on a capacitor while charging is given by:

$Q=C{V}_b(1-e^{-\frac{t}{R_{c}C}})$

where $V_b$ is the voltage of the battery, $R_c$ is the total resistance in the charging circuit, t is the time since the capacitor started to charge and C is the capacitance.

The equation for the charge on the capacitor while discharging is:

$Q=Q_0{e}^{-\frac{t}{R_{d}C}}$

where $Q_0$ is the charge on the capacitor just before it starts to discharge, $R_d$ is the total resistance in the discharging circuit and $t$ now represents the time since the capacitor started to discharge.

$V_b$