Do the RC formulas apply to more complex networks?

The formulas apply exactly when R and C form a simple series loop with the battery. For complex networks, use Kirchhoff's rules to reduce to equivalent R and C. The exponential form then holds with τ = R_eq times C_eq.

Charge & discharge of a condenser

Q: Why is the time constant τ = RC?

From Kirchhoff's voltage rule in the charging circuit and using i = dq/dt, we get a differential equation whose solution contains an exponential term e to the minus t over RC. The quantity RC has units of time and controls how quickly the exponential decays.

Q: What does the time constant physically mean?

τ = RC is the characteristic time of the RC circuit - the time for the capacitor to reach 63% of its final charge during charging, or drop to 37% during discharging. Small τ means quick response; large τ means sluggish response.

Q: How long does it take for a capacitor to charge fully?

Mathematically, a capacitor takes infinite time to reach exactly the maximum charge. Practically, after 5τ it reaches about 99.3%, and after 10τ over 99.99%. Engineers treat 5τ as fully charged.

Q: Why does the current in an RC charging circuit start at maximum and decrease?

At t = 0, the uncharged capacitor acts like a short circuit, so full battery voltage drops across the resistor, giving maximum current V₀/R. As charge accumulates, the capacitor's voltage opposes the battery, reducing net driving voltage and hence the current.

Q: In a discharging capacitor, where does the stored energy go?

All of it dissipates as heat in the resistor. Integrating power dissipation i squared R over time equals the initial stored energy q₀ squared over 2C. If other components are present, some energy may temporarily go elsewhere.

Q: Can the time constant be changed without changing R or C?

Not for a single-loop RC circuit. But adding resistors or capacitors changes the effective R and C. Inserting a dielectric between plates increases C and hence τ without changing R.

PhysicsCapacitorsFor JEE aspirants

When a capacitor is connected to a battery through a resistor, its charge does not jump instantly to the maximum value - it rises exponentially over time, controlled by the time constant $τ = R C$ . Similarly, when a charged capacitor discharges through a resistor, its charge falls exponentially. Understanding the RC charging and discharging curves is essential for JEE and NEET Physics questions on transient currents, exponential decay, and RC circuit analysis.

Key Formulas - Quick Reference

Charging: $q (t) = q_{0} [1 - e^{- t / R C}]$
Discharging: $q (t) = q_{0} e^{- t / R C}$
Time constant: $τ = R C$ (SI unit: seconds)
Charging voltage: $V (t) = V_{0} [1 - e^{- t / R C}]$
Discharging voltage: $V (t) = V_{0} e^{- t / R C}$
Transient current (charging): $i (t) = i_{0} e^{- t / R C}$ where $i_{0} = V_{0} / R$
Transient current (discharging): $i (t) = - i_{0} e^{- t / R C}$ (opposite direction)
At $t = τ$ : charging reaches 63% of $q_{0}$ ; discharging drops to 37% of $q_{0}$

1. Charging a Capacitor Through a Resistor

Figure: RC circuit with two-position switch. Position 1 charges C through R from V₀; position 2 discharges C through R via the top wire. Both use time constant τ = RC.

Consider a battery of EMF $ε$ connected through a resistor $R$ to a capacitor $C$ via a switch. When the switch is closed at $t = 0$ , current begins to flow and the capacitor starts charging.

The capacitor takes a finite time to charge - it does not jump instantly to full charge. As charge accumulates on the plates, the growing voltage across the capacitor opposes the battery, so the current gradually decreases.

Charge as a function of time

q (t) = q_{0} [1 - e^{- t / R C}]

where:

$q_{0} = C ε$ is the final maximum charge (at $t \to \infty$ )
$R$ is the total resistance in the charging circuit
$C$ is the capacitance

Figure: Charging curve

q (t) = q_{0} (1 - e^{(- t / R C)})

. At t = τ = RC, q = 0.63 q₀.

The time constant

The quantity $τ = R C$ is called the time constant of the RC circuit. It has units of seconds. Setting $t = τ = R C$ in the charging equation:

q (τ) = q_{0} [1 - e^{- 1}] = q_{0} [1 - \frac{1}{e}] = q_{0} (1 - 0.37) = 0.63 q_{0}

The time constant

τ = R C

is the time in which:

the charge on the capacitor rises to 63% of its maximum value (during charging), or
the charge falls to 37% of its initial value (during discharging).

Voltage across the capacitor during charging

Since $V (t) = q (t) / C$ , the voltage follows the same exponential curve:

V (t) = V_{0} [1 - e^{- t / R C}] where V_{0} = ε

Transient current during charging

The current in the circuit is largest at $t = 0$ (when the capacitor is empty and acts like a short circuit) and decays exponentially to zero:

i (t) = i_{0} e^{- t / R C} where i_{0} = \frac{V _{0}}{R}

At $t = τ = R C$ , the current has fallen to $i_{0} / e = 0.37 i_{0}$ , or 37% of its initial value.

2. Discharging a Capacitor Through a Resistor

If a fully charged capacitor (initial charge $q_{0}$ ) is connected directly across a resistor $R$ (no battery), it discharges through the resistor.

Charge decay

q (t) = q_{0} e^{- t / R C}

Figure: Discharging curve

q (t) = q_{0} e^{(- t / R C)}

. At t = τ = RC, q = 0.37 q₀.

At $t = τ = R C$ :

q (τ) = \frac{q _{0}}{e} = 0.37 q_{0} (37% of q_{0})

Voltage and current during discharge

Voltage across the capacitor:

V (t) = V_{0} e^{- t / R C}

Transient discharge current (note the negative sign - direction is opposite to charging):

i (t) = - i_{0} e^{- t / R C}

Dimensional check on the time constant

$R$ has dimensions $[M L^{2} T^{- 3} A^{- 2}]$ and $C$ has dimensions $[M^{- 1} L^{- 2} T^{4} A^{2}]$ . So:

[R C] = [M L^{2} T^{- 3} A^{- 2}] \times [M^{- 1} L^{- 2} T^{4} A^{2}] = [T]

The product $R C$ is indeed a time, confirming its interpretation as the time constant. Correspondingly, $1/ (R C)$ has dimensions of frequency $[T^{- 1}]$ .

3. Comparison of Charging and Discharging

Quantity	Charging	Discharging
Charge $q (t)$	$q_{0} (1 - e^{- t / R C})$ , rising	$q_{0} e^{- t / R C}$ , falling
At $t = τ = R C$	63% of $q_{0}$	37% of $q_{0}$
At $t = 5 τ$	99.3% of $q_{0}$ (nearly full)	0.7% of $q_{0}$ (nearly empty)
Current direction	Positive (flowing into capacitor)	Reversed (flowing out of capacitor)
Initial current magnitude	$i_{0} = V_{0} / R$	$i_{0} = V_{0} / R$

Rule of thumb: After a time of about

5 τ

, a capacitor is essentially fully charged (or fully discharged) - it reaches 99% of its final value. This is often taken as "practically complete" in engineering estimates.

Solved Examples

Solved Example 1

10 μ F

capacitor is charged through a

2 M Ω

resistor by a

12 V

battery. Find (a) the time constant, (b) the time taken to reach

6 V

across the capacitor, and (c) the current in the circuit at

t = 0

Solution:

(a) Time constant:

τ = R C = (2 \times 1 0^{6}) (10 \times 1 0^{- 6}) = 20 s

(b) Time to reach $6 V$ . This is exactly half the final voltage:

V (t) = 12 [1 - e^{- t /20}] = 6 \Rightarrow e^{- t /20} = \frac{1}{2}

\Rightarrow \frac{t}{20} = ln 2 \approx 0.693 \Rightarrow t \approx 13.9 s

(c) Initial current (at $t = 0$ , capacitor is uncharged and acts like a short):

i_{0} = \frac{V _{0}}{R} = \frac{12}{2 \times 1 0 ^{6}} = 6 \times 1 0^{- 6} A = 6 μ A

Solved Example 2

A fully charged

5 μ F

capacitor holds

100 μ C

of charge. It is then discharged through a

50 k Ω

resistor. Find (a) the initial voltage, (b) the time constant, (c) the charge remaining after

0.5 s

Solution:

(a) Initial voltage:

V_{0} = \frac{q _{0}}{C} = \frac{100 \times 1 0 ^{- 6}}{5 \times 1 0 ^{- 6}} = 20 V

(b) Time constant:

τ = R C = (50 \times 1 0^{3}) (5 \times 1 0^{- 6}) = 0.25 s

(c) Charge at $t = 0.5 s$ :

t / τ = 0.5/0.25 = 2

q (0.5) = 100 e^{- 2} \approx 100 \times 0.135 = 13.5 μ C

Solved Example 3

In an RC charging circuit, the charge on the capacitor reaches 87% of its final value in

4 s

. Find the time constant

τ

Solution:

Using the charging equation:

0.87 q_{0} = q_{0} [1 - e^{- t / τ}] \Rightarrow e^{- 4/ τ} = 0.13

Taking natural log:

- \frac{4}{τ} = ln (0.13) \approx - 2.04

τ = \frac{4}{2.04} \approx 1.96 s

Answer: $τ \approx 2 s$ .

Common Mistakes to Avoid

Watch out

Mixing up charging and discharging equations. Charging has the $(1 - e^{- t / R C})$ factor (rising to $q_{0}$ ); discharging has just $e^{- t / R C}$ (falling from $q_{0}$ ).
Forgetting that current direction reverses during discharge. Charging current flows into the capacitor from the battery; discharging current flows out of the capacitor through the resistor - opposite direction.
Assuming a capacitor charges instantly. Only in the limit of zero resistance ( $R \to 0$ ) does charging become instantaneous. Any real circuit has some resistance, so the process takes about $5 τ$ to be "practically complete."
Using $e = 2.72$ (rounding too much). The correct value is $e \approx 2.718$ , so $1/ e \approx 0.368$ (not $0.37$ exactly). For quick calculations $0.37$ is acceptable, but derive precisely for MCQs testing decimal accuracy.
Confusing the time constant with the time to full charge. $τ$ is when the capacitor reaches only 63% of maximum charge. It takes about $5 τ$ to be "fully" charged (99%).

Frequently Asked Questions

Q1. Why is the time constant $τ = R C$ ?

From Kirchhoff's voltage rule in the charging circuit, $ε = i R + q / C$ , and using $i = d q / d t$ , we get the differential equation $R d q / d t + q / C = ε$ . Its solution has an exponential term $e^{- t / (R C)}$ . The quantity $R C$ has units of time and controls how quickly the exponential decays.

Q2. What does the time constant physically mean?

$τ = R C$ is the "characteristic time" of the RC circuit - roughly, the time for the capacitor to reach 63% of its final charge during charging, or drop to 37% during discharging. If $τ$ is small, the circuit reacts quickly to voltage changes; if $τ$ is large, the circuit responds sluggishly.

Q3. How long does it take for a capacitor to charge fully?

Mathematically, a capacitor takes infinite time to charge to exactly $q_{0}$ (the exponential approaches asymptotically). Practically, after $5 τ$ it reaches about 99.3% of $q_{0}$ , and after $10 τ$ over 99.99%. Engineers typically treat $5 τ$ as "fully charged."

Q4. Why does the current in an RC charging circuit start at maximum and decrease?

At $t = 0$ , the uncharged capacitor has zero voltage across it, so it acts like a short circuit and the full battery voltage drops across the resistor, giving maximum current $V_{0} / R$ . As charge accumulates, the capacitor's own voltage grows and opposes the battery, reducing the net driving voltage across the resistor - and hence the current.

Q5. In a discharging capacitor, where does the stored energy go?

All of it dissipates as heat in the resistor. The initial energy stored is $U_{0} = q_{0}^{2} / (2 C)$ ; integrating the power dissipation $i^{2} R$ over all time gives exactly this amount. If the discharge circuit contains other components (like an inductor), some energy may be temporarily stored elsewhere.

Q6. Can the time constant be changed without changing $R$ or $C$ ?

Not for a single-loop RC circuit. However, if you have multiple capacitors and resistors, adding them changes the effective $R$ and $C$ seen by the circuit. Inserting a dielectric between the plates increases $C$ (and hence $τ$ ) without changing $R$ .

Q7. How does the RC time constant relate to signal filtering in electronics?

The RC circuit is the basis of first-order high-pass and low-pass filters. The cutoff frequency (where signals are attenuated by a factor of $2$ ) is $f_{c} = 1/ (2 π R C)$ . Larger $τ$ means lower cutoff, blocking higher frequencies - useful in noise filtering and signal shaping.

Q8. Does the RC formula apply if the resistor is in series with the capacitor differently, or is it always this simple form?

The formulas apply exactly when $R$ and $C$ form a simple series loop with the battery. For more complex networks (multiple resistors, capacitors, batteries), use Kirchhoff's rules to reduce the network. The equivalent $R$ and $C$ then give the same exponential form with $τ = R_{e q} C_{e q}$ .

Previous year questions on Charge & discharge of a condenser

10 questions from past papers, each with a step-by-step solution.

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1. Charging a Capacitor Through a Resistor

Charge as a function of time

The time constant

Voltage across the capacitor during charging

Transient current during charging

2. Discharging a Capacitor Through a Resistor

Charge decay

Voltage and current during discharge

Dimensional check on the time constant

3. Comparison of Charging and Discharging

Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why is the time constant τ=RC?

Q2. What does the time constant physically mean?

Q3. How long does it take for a capacitor to charge fully?

Q4. Why does the current in an RC charging circuit start at maximum and decrease?

Q5. In a discharging capacitor, where does the stored energy go?

Q6. Can the time constant be changed without changing R or C?

Q7. How does the RC time constant relate to signal filtering in electronics?

Q8. Does the RC formula apply if the resistor is in series with the capacitor differently, or is it always this simple form?

Previous year questions on Charge & discharge of a condenser

Ready to master Capacitors?

Q1. Why is the time constant $τ = R C$ ?

Q6. Can the time constant be changed without changing $R$ or $C$ ?