A capacitor is connected to a 20 V battery through a resistance of 10$$\Omega$$. It is found that the potential difference across the capacitor rises to 2 V in 1 $$\mu$$s. The capacitance of the capacitor is _________ $$\mu$$F.
Given $$\ln\frac{10}{9} = 0.105$$

For a series combination of a resistor $$R$$ and a capacitor $$C$$ connected to a constant supply voltage $$V$$, the rise of voltage across the capacitor with time is governed by the charging equation

$$V_C(t)=V\bigl(1-e^{-t/RC}\bigr).$$

Here we have a battery of $$V=20\ \text{V}$$, a resistance $$R=10\ \Omega$$ and an observed capacitor voltage $$V_C=2\ \text{V}$$ after a time interval $$t=1\ \mu\text{s}=1\times10^{-6}\ \text{s}.$$ Substituting these numbers into the formula gives

$$2 = 20\left(1-e^{-t/RC}\right).$$

Dividing both sides by $$20$$, we obtain

$$\frac{2}{20}=1-e^{-t/RC}.$$

This simplifies to

$$0.1 = 1-e^{-t/RC}.$$

Re-arranging, we isolate the exponential term:

$$e^{-t/RC}=1-0.1=0.9.$$

Taking the natural logarithm of both sides, we write

$$-\,\frac{t}{RC}=\ln(0.9).$$

Since $$\ln(0.9)=-\ln\!\left(\dfrac{10}{9}\right)$$, we have

$$-\,\frac{t}{RC}= -\,\ln\!\left(\dfrac{10}{9}\right).$$

Removing the minus signs on both sides gives

$$\frac{t}{RC}= \ln\!\left(\dfrac{10}{9}\right).$$

We now substitute the known numerical values: $$t = 1\times10^{-6}\ \text{s}$$, $$R = 10\ \Omega$$ and, from the question, $$\ln\!\left(\dfrac{10}{9}\right)=0.105.$$ Thus,

$$\frac{1\times10^{-6}}{10\,C}=0.105.$$

Multiplying both sides by $$10C$$, we obtain

$$1\times10^{-6}=1.05\,C.$$

Dividing by $$1.05$$ isolates the capacitance:

$$C=\frac{1\times10^{-6}}{1.05}.$$

Carrying out the division,

$$C\approx0.952\times10^{-6}\ \text{F}.$$

Since $$1\times10^{-6}\ \text{F}=1\ \mu\text{F}$$, we convert the answer to microfarads:

$$C\approx0.952\ \mu\text{F}\,.$$

Rounding to two significant figures, this is $$0.95\ \mu\text{F}$$.

Hence, the correct answer is Option A.