A capacitor is connected to a 20 V battery through a resistance of 10 . It is found that the potential difference across the capacitor rises to 2 V in 1 s. The capacitance of the capacitor is __________ F. Given :

Key concepts you must know

1. RC Charging Law
   When a capacitor of capacitance $C$ is charged through a resistor $R$ by a battery of emf $V_0$, its voltage at any time $t$ is
   $V_C(t) = V_0 \left(1 - e^{-t/RC}\right)$ Here $RC$ is called the time constant (symbol $\tau$). After time $\tau$, the capacitor reaches about 63 % of the final voltage.

2. Time Constant ($\tau$)
   $\tau = R C$
   It tells how quickly the capacitor charges. Smaller $R$ or $C$ → smaller $\tau$ → faster charging.

3. Natural Logarithm in rearranging
   To extract $C$, you will take natural logs by isolating the exponential term.

Logical chain to attack the problem

1. Write the charging formula with the given numbers ($V_0 = 20\,\text{V}$, $V_C = 2\,\text{V}$, $t = 1\,\text{s}$, $R = 10\,\Omega$).
2. Rearrange to isolate the exponential: find $e^{-t/RC}$.
3. Take ln (natural log) on both sides to solve for $RC$.
4. Divide by the known resistance to obtain $C$.
5. Quote units properly in farads (F).

Following these disciplined steps ensures you avoid algebra slips and remember the physical meaning of every quantity.

Imagine filling a bottle with water through a narrow pipe

• Battery is like a water tank kept 20 meters high – it pushes water down with 20-V worth of pressure.
• Resistor is a skinny pipe that slows the flow, labelled 10 ohms (like a 10-point speed-breaker).
• Capacitor is the bottle we are filling. Its voltage tells us how much water is already inside.

At the very start the bottle is empty (0 V). After 1 second we notice the bottle’s voltage has climbed to 2 V.

The question: How big is this bottle (its capacitance) if it rises that slowly?

Bigger bottles fill more slowly, so if it took a whole second just to reach 2 V (only one-tenth of the supply), the bottle must be quite large. We will use the special charging rule for capacitors to work out exactly how large in farads.

Step-by-step calculation

1. Start with charging equation

   $$V_C(t) = V_0 \left(1 - e^{-t/RC}\right)$$

2. Insert given values

   $$2 = 20 \left(1 - e^{-1/(R C)}\right)$$

3. Isolate the exponential term

   $$\frac{2}{20} = 1 - e^{-1/(RC)} \quad \Longrightarrow\quad 0.1 = 1 - e^{-1/(RC)}$$ $$e^{-1/(RC)} = 1 - 0.1 = 0.9$$

4. Take natural logarithm

   $$-\frac{1}{RC} = \ln(0.9)$$ $$\ln(0.9) \approx -0.1053605$$ $$\therefore \frac{1}{RC} = 0.1053605$$

5. Find the time constant $RC$

   $$RC = \frac{1}{0.1053605} \approx 9.495\,\text{s}$$

6. Extract capacitance (given $R = 10\,\Omega$)

   $$C = \frac{RC}{R} = \frac{9.495}{10} \approx 0.9495\,\text{F}$$

7. Final answer

   $$\boxed{C \;\approx\; 0.95\,\text{F}}$$

Thus the capacitor’s capacitance is roughly 0.95 farads.