Answer
$6.1K\Omega$
Work Step by Step
The required resistance can be determined as follows:
$q=\mathcal{E}C(1-e^{\frac{-t}{RC}})$
We plug in the known values to obtain:
$0.9\mathcal{E}C=\mathcal{E}C(1-e^{\frac{-21s}{R\times 1500\times 10^{-6}F}})$
$\implies e^{\frac{-21s}{R\times 1500\times 10^{-6}F}}=1-0.9$
$e^{\frac{-21s}{R\times 1500\times 10^{-6}F}}=0.1$
Taking log on both sides, we obtain:
${\frac{-21s}{R\times 1500\times 10^{-6}F}}=ln(0.1)$
${\frac{-21s}{R\times 1500\times 10^{-6}F}}=-2.30$
$\implies R=\frac{21s}{2.30\times 1500\times 10^{-6}F}$
$R=6.08\times 10^3\Omega$
$R=6.1K\Omega$
