RC-Circuit

PhysicsCurrent ElectricityFor JEE aspirants

An RC circuit consists of a resistor and a capacitor in series with a DC source or switch. When the switch is closed, the capacitor charges through the resistor, and the charge grows exponentially as $q (t) = q_{m a x} (1 - e^{- t / R C})$ . When the source is removed and the capacitor discharges through the resistor, the charge decays as $q (t) = q_{0} e^{- t / R C}$ . The product $τ = R C$ is the time constant, representing the time in which charge reaches $63%$ of its final value during charging (or drops to $37%$ during discharge). RC circuits appear in JEE Physics - Current Electricity and NEET Physics - Current Electricity, and form the basis for timers, filters, and coupling circuits.

Key Formulas - Quick Reference

Time constant: $τ = R C$ (units: seconds)
Charging: $q (t) = q_{m a x} (1 - e^{- t / R C})$ , $q_{m a x} = εC$
Charging current: $i (t) = \frac{ε}{R} e^{- t / R C} = i_{0} e^{- t / R C}$
Discharging: $q (t) = q_{0} e^{- t / R C}$
Discharging current: $i (t) = - \frac{q _{0}}{R C} e^{- t / R C} = - i_{0} e^{- t / R C}$
Voltage across capacitor (charging): $V_{C} = ε (1 - e^{- t / R C})$
Voltage across capacitor (discharging): $V_{C} = ε e^{- t / R C}$
Energy stored at full charge: $U = \frac{1}{2} C ε^{2}$
Heat dissipated during full charging: $H = \frac{1}{2} C ε^{2}$ (equals stored energy)

1. RC Circuit - Charging

Consider a series RC circuit: a battery of EMF $ε$ , a resistor $R$ , an uncharged capacitor $C$ , and a switch $S$ . For $t < 0$ , the switch is open and $q = 0$ . At $t = 0$ , the switch is closed.

Figure 1: RC charging circuit - closing switch

S

t = 0

starts a decaying current

i (t) = (ε / R) e^{- t / R C}

that charges capacitor

C

through resistor

R

Derivation

At time $t$ , let $q$ be the charge on the capacitor. Apply Kirchhoff's loop rule (going around the loop from the negative terminal):

ε - i R - \frac{q}{C} = 0

Since $i = d q / d t$ :

R \frac{d q}{d t} = ε - \frac{q}{C} \Rightarrow R C \frac{d q}{d t} = εC - q

Separating variables:

\frac{d q}{εC - q} = \frac{d t}{R C}

Integrating from $q = 0$ at $t = 0$ to $q$ at $t$ :

- ln ∣ εC - q ∣_{0}^{q} = \frac{t}{R C}

ln \frac{εC - q}{εC} = - \frac{t}{R C}

\frac{εC - q}{εC} = e^{- t / R C}

q (t) = εC (1 - e^{- t / R C}) = q_{m a x} (1 - e^{- t / R C})

where $q_{m a x} = εC$ is the final (fully-charged) charge.

Current during charging:

i (t) = \frac{d q}{d t} = \frac{εC}{R C} e^{- t / R C} = \frac{ε}{R} e^{- t / R C} = i_{0} e^{- t / R C}

where $i_{0} = ε / R$ is the initial current (as if the capacitor were absent).

Voltage across capacitor:

V_{C} = \frac{q}{C} = ε (1 - e^{- t / R C})

2. Time Constant $τ = R C$

The quantity $τ = R C$ is called the time constant of the circuit. It has units of seconds ( $Ω \cdot F$ = s) and represents the characteristic time scale of exponential build-up or decay.

Physical meaning:

At $t = τ$ (charging): $q = q_{m a x} (1 - e^{- 1}) = q_{m a x} (1 - 0.368) \approx 0.63 q_{m a x}$ . Charge reaches $63%$ of its final value.
At $t = 2 τ$ : $q \approx 0.865 q_{m a x}$ .
At $t = 5 τ$ : $q \approx 0.993 q_{m a x}$ - effectively fully charged.
At $t = τ$ (discharging): $q$ drops to $37%$ of initial.

3. RC Circuit - Discharging

Now suppose the capacitor has charge $q_{0}$ at $t < 0$ , and at $t = 0$ the battery is removed (or the switch is flipped to short the capacitor through $R$ ). The capacitor discharges through the resistor.

Derivation

Loop rule (no source):

\frac{q}{C} + i R = 0 \Rightarrow R \frac{d q}{d t} = - \frac{q}{C}

Separating and integrating from $q_{0}$ at $t = 0$ to $q$ at $t$ :

\int_{q_{0}}^{q} \frac{d q}{q} = - \frac{1}{R C} \int_{0}^{t} d t

ln \frac{q}{q _{0}} = - \frac{t}{R C}

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

Current during discharging:

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

The negative sign indicates that the discharging current flows in a direction opposite to the charging current - the capacitor now acts like a source.

Voltage across capacitor:

V_{C} (t) = \frac{q}{C} = \frac{q _{0}}{C} e^{- t / R C} = ε e^{- t / R C}

Figure 2: (Left) Charging curve - charge builds up to

63%

of maximum at

t = a u = R C

. (Right) Discharging curve - charge falls to

37%

of initial at

t = a u

4. Energy Considerations

During complete charging from $q = 0$ to $q = εC$ :

Total energy supplied by battery: $W_{batt} = εQ = ε (εC) = C ε^{2}$ .
Energy stored in capacitor: $U_{C} = \frac{1}{2} C ε^{2}$ .
Energy dissipated in resistor: $H = W_{batt} - U_{C} = \frac{1}{2} C ε^{2}$ .

Notable result: during full charging, exactly half the battery energy is dissipated as heat in the resistor, independent of $R$ - the other half is stored in the capacitor. Efficiency of charging is 50%.

Solved Example 1

Q: Calculate the steady-state current in the $2 Ω$ resistor of the circuit. Battery has EMF $6$ V and negligible internal resistance; there is a $4 Ω$ resistor in series with a $0.2 μ$ F capacitor, and this series branch is in parallel with a $3 Ω$ resistor. This parallel combination is in series with a $2.8 Ω$ resistor and then the $2 Ω$ resistor across the battery.

Solution

Steady state: no current flows through the capacitor (it is fully charged), and hence no current flows through the $4 Ω$ resistor in series with the capacitor.

Effective resistance of the parallel combination of $2 Ω$ (the branch we want) and $3 Ω$ :

\frac{1}{R _{p}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \Rightarrow R_{p} = 1.2 Ω

Total resistance: $R_{total} = 2.8 + 1.2 = 4 Ω$ .

Total current from battery: $I = 6/4 = 1.5$ A.

Voltage across the parallel combination: $V_{A B} = 1.5 \times 1.2 = 1.8$ V.

Current through the $2 Ω$ resistor: $I_{2} = 1.8/2 = 0.9$ A.

Solved Example 2

Q: A $10 μ$ F capacitor is charged to $100$ V and then discharged through a $50 k Ω$ resistor. (a) Find the time constant. (b) Find the voltage across the capacitor after $0.5$ s. (c) After how long does the voltage drop to $10$ V?

Solution

(a) $τ = R C = (5 \times 1 0^{4}) (10 \times 1 0^{- 6}) = 0.5$ s.

(b) At $t = 0.5$ s $= τ$ :

V (0.5) = 100 e^{- 1} = 100 \times 0.368 = 36.8 V

(c) $V (t) = 10 \Rightarrow 100 e^{- t /0.5} = 10 \Rightarrow e^{- t /0.5} = 0.1$

- \frac{t}{0.5} = ln (0.1) = - 2.303 \Rightarrow t = 1.15 s

Practical Applications

RC circuits form the basis of many devices: wiper-blade delay timers (large $τ$ ), camera flash storage (large $C$ , quick discharge), signal filters (frequency-selective response depends on $τ$ ), and clock circuits in digital electronics. The concept of exponential decay with time constant $τ$ also appears in radioactive decay, LR circuits, and cooling of bodies (Newton's law of cooling).

Frequently Asked Questions

Why is the charging current maximum at

t = 0

and zero at

t = \infty

? (JEE / NEET)

t = 0

, the capacitor has no charge, so there is no back-voltage; the full EMF

ε

drives current

i_{0} = ε / R

through the resistor. As charge accumulates, the capacitor voltage

q / C

opposes

ε

, reducing the net driving voltage and hence the current. When fully charged,

q / C = ε

, net voltage across

R

is zero, and current stops.

What is the physical meaning of the time constant

τ = R C

? (JEE Main / NEET)

The time constant

τ = R C

is the time in which the charge on a charging capacitor reaches

63%

(i.e.

1 - 1/ e

) of its final value, or in which a discharging capacitor drops to

37%

(i.e.

1/ e

) of its initial value. Physically, larger

R

slows current flow, larger

C

requires more charge to reach a given voltage - both increase the time scale of the process.

Why is only half the battery energy stored in the capacitor during charging? (JEE Advanced-level insight)

Independent of the value of

R

, exactly

\frac{1}{2} C ε^{2}

is dissipated as heat during the charging process, while the same amount is stored in the capacitor. Total energy supplied by the battery is

C ε^{2}

. This 50-50 split arises because the average voltage across the capacitor during charging is

ε /2

, so the average charge

\times

voltage delivered by the battery gives half the energy to the capacitor and half to the resistor as heat.

Does a capacitor allow DC current in steady state? (JEE Main / NEET)

No. In steady state, once the capacitor is fully charged to the applied voltage, no current flows through it - it behaves as an open circuit for DC. This is why problems ask for steady-state currents by ignoring any branch containing only a capacitor. For AC signals, however, the capacitor continuously charges and discharges, so it does allow AC current with impedance

1/ (ω C)

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Practice questions →Take a mock test

1. RC Circuit - Charging

Derivation

2. Time Constant τ=RC

3. RC Circuit - Discharging

Derivation

4. Energy Considerations

Frequently Asked Questions

Ready to master Current Electricity?

2. Time Constant $τ = R C$