RC circuits — charging, discharging, time constant τ = RC: Question bank

1.8.19 · D5 · Physics › Electromagnetism › RC circuits — charging, discharging, time constant τ = RC

Definition Is page ke paanch symbols (har ek earned hai)

Neeche sab kuch inhi ko reuse karta hai — baad mein koi nayi notation nahi aati.

$R$ — resistance (ohms mein, $Ω$ ): resistor charge ke flow ko kitna rokta hai. Bada $R$ = patli "straw." Dekho Ohm's Law and Resistance.
$C$ — capacitance (farads mein, F): capacitor har volt par kitna charge hold karta hai. Bada $C$ = chauda "bucket." Dekho Capacitors and Capacitance.
$Q$ — charge (coulombs mein, C) jo abhi capacitor ki plates par baitha hai. Charging ke waqt badhta hai, discharging mein ghatta hai.
$t$ — time (seconds mein, s), us moment se measure hota hai jab switch flip hota hai $t = 0$ par.
$τ = R C$ — time constant jo upar ke dono se banta hai: seconds mein ek time jo set karta hai ki sab kuch kitni tezi se hoga.

Do aur jo hum parent se lete hain: $E$ battery ki emf hai (uski push, volts mein), aur $V_{C} = Q / C$ capacitor ki apni voltage hai, jo badhti hai jab $Q$ ikattha hota hai aur battery ke khilaf fight back karta hai. Current charge ka flow hai, $I = \frac{d Q}{d t}$ .

Neeche ke do figures woh mental pictures hain jinpar har question tika hai — circuit aur do curves. Jab bhi koi trap slippery lage, inhe dobara dekh lo.

Figure — RC circuits — charging, discharging, time constant τ = RC

True or false — justify

A capacitor charges at a constant rate because the battery voltage is constant.

False. Current set hota hai net voltage

E - V_{C}

se divided by

R

; jab

V_{C}

badhta hai toh net push kam hoti hai, isliye rate girta hai — yahi exactly woh cheez hai jo curve ko exponential banati hai, straight line nahi.

t = τ

a charging capacitor is essentially full.

False. Ek

τ

par yeh sirf

1 - e^{- 1} \approx 63.2%

tak pahuncha hota hai. "Essentially full" (>99%) lagbhag

5 τ

laita hai, kyunki

e^{- 5} \approx 0.7%

baki rehta hai.

Charging reaches 63% and discharging falls to 63% — same number, same

τ

False. Charging

1 - e^{- 1} \approx 63.2%

tak upar jaati hai; discharging

e^{- 1} \approx 36.8%

tak neeche aati hai. Yeh complementary hain (

63.2% + 36.8% = 100%

), identical nahi.

Doubling both

R

and

C

makes the circuit charge four times slower.

True timescale ke hisaab se.

τ = R C

ban jaata hai

4 τ

, toh har stage chaar guna zyada time leti hai. Curve ki shape unchanged rehti hai — sirf uski horizontal stretch badlti hai.

The charging current and discharging current both decay with the same time constant.

True. Dono obey karte hain

I \propto e^{- t / τ}

τ = R C

ke saath; resistor aur capacitor same hain, isliye exponential clock dono directions mein same hai.

A larger capacitor always charges to a higher final voltage.

False. Final voltage

E

hoti hai (battery ki push),

C

se independent. Bada

C

sirf zyada charge store karta hai (

Q_{m a x} = E C

) aur wahan pahunchne mein zyada time leta hai.

During discharge the current flows in the same direction as during charging.

False. Charging ke waqt, current positive plate ke andar jaata hai; discharging ke waqt, capacitor current bahar usi plate se drive karta hai, isliye

R

ke through current direction reverse ho jaati hai.

At the very instant charging begins, the resistor carries the maximum possible current.

True.

t = 0

par,

V_{C} = 0

, toh poora emf

R

par baitha hai aur

I_{0} = E / R

— sabse bada current jo loop kabhi dekhta hai.

Energy delivered by the battery while charging equals the energy stored in the capacitor.

False. Battery ki energy ka exactly aadha

R

mein heat ke roop mein dissipate hota hai; sirf doosra aadha capacitor mein stored hota hai —

R

ki value se independent. Dekho Energy Stored in a Capacitor.

Spot the error

"At

t = 0

the capacitor is empty, so its voltage is

E

Error cause aur effect ko swap karta hai. Empty matlab

Q = 0

, toh

V_{C} = Q / C = 0

; poora voltage

E

resistor par baitha hai, capacitor par nahi,

t = 0

par.

"To find when charging is 90% done, I set

e^{- t / τ} = 0.90

Wrong quantity. Charging fraction hai

1 - e^{- t / τ}

, toh 90% done matlab

1 - e^{- t / τ} = 0.90

, yaani

e^{- t / τ} = 0.10

e^{- t / τ} = 0.90

set karna woh waqt dhundta hai jab yeh sirf 10% done hai.

"Since

τ = R C

has units of

Ω \cdot F

, and those aren't seconds,

τ

isn't really a time."

Units seconds mein collapse ho jaati hain:

Ω \cdot F = \frac{V}{A} \cdot \frac{C}{V} = \frac{C}{A} = \frac{C}{C / s} = s

. Mismatch sirf surface-level notation ki baat hai.

"During discharge

d Q / d t = + Q / (R C)

, so charge grows."

Sign error. KVL with no source deta hai

R \frac{d Q}{d t} = - \frac{Q}{C}

, toh

\frac{d Q}{d t} = - \frac{Q}{R C} < 0

— charge decay karta hai. Minus sign encode karta hai "charge kho raha hai."

"A capacitor blocks DC, so no current ever flows when I connect the battery."

Yeh DC ko sirf final steady state mein block karta hai. Transient ke dauran (roughly pehle kuch

τ

), current zaroor bahta hai charge deliver karne ke liye; yeh zero ho jaata hai tabhi jab

V_{C} = E

"Because current stops at the end, the resistor was useless — remove it and it charges instantly."

R \to 0

ke saath,

τ = R C \to 0

, toh initial current spike

E / R \to \infty

. Reality mein kuch bhi "instantly" charge nahi hota; resistor hi woh cheez hai jo current ko finite banata hai aur process ko well-behaved rakhta hai.

Why questions

Why does the charging curve bend over (concave down) instead of being a straight line?

Kyunki rate

\frac{d Q}{d t} = \frac{1}{R C} (Q_{m a x} - Q)

proportional hai abhi kitna jaana baki hai us par. Jab

Q

Q_{m a x}

ke paas aata hai toh woh gap shrink hota hai, isliye slope continuously flat hoti jaati hai.

Why do charging and discharging share the exact same

τ

Time constant sirf loop ke components par depend karta hai,

R

aur

C

par, battery ki presence par nahi. Dono processes same

R C

product se governed hote hain, isliye dono same clock par chalte hain.

Why is the current identical in magnitude everywhere in a series RC circuit at any instant?

Series matlab ek path bina junctions ke, aur current charge ka flow hai — charge wire mein ikattha nahi ho sakta, isliye same

I = d Q / d t

R

se guzarta hai aur capacitor plate par simultaneously aata hai. Yeh Kirchhoff's ka companion hai, current continuity.

Why does the same mathematics describe Newton's Law of Cooling and radioactive decay?

Teeno obey karte hain "rate of change proportional to how far from equilibrium," jo ek solution force karta hai jo

e^{- t / τ}

se bana hota hai. Dekho Exponential Decay and Differential Equations — physics alag hai lekin equation ek hi template hai.

Why is

τ = L / R

for an LR circuit but

τ = R C

here?

RC circuit mein bada

R

cheezein slow karta hai (kam current), isliye

R

multiply karta hai. LR circuit mein bada

R

current ke decay ko speed up karta hai, isliye

R

divide karta hai.

R

ka role flip hota hai kyunki ek mein current limit karta hai aur doosre mein stored current-energy dissipate karta hai.

Why does a bigger

R

make charging slower even though it lowers the current the whole time?

Slower charging matlab same total charge

Q_{m a x} = E C

deliver karne mein zyada waqt lagta hai. Bada

R

har instant par delivery rate throttle karta hai, toh us fixed charge ko accumulate karne mein simply zyada time lagta hai.

Edge cases

What happens as

R \to 0

(ideal wire, no resistor)?

τ = R C \to 0

, toh charging effectively instantaneous hai aur initial current

E / R \to \infty

— ek idealized infinite spike. Real circuits mein hamesha kuch resistance hoti hai jo ise tame karti hai.

What happens as

R \to \infty

(broken wire / open switch)?

τ = R C \to \infty

, toh current

E / R \to 0

aur koi charge kabhi nahi bahta — capacitor jis bhi voltage par tha wahan hamesha ke liye reh jaata hai. Infinite resistance bas ek open circuit hai.

What happens as

C \to 0

(vanishingly small capacitor)?

τ \to 0

aur

Q_{m a x} = E C \to 0

: yeh essentially zero charge par "instantly charge" ho jaata hai. Zero-capacitance element ek aisi open gap ki tarah behave karta hai jo kabhi kuch store nahi karta.

What happens as

C \to \infty

(enormous capacitor)?

τ \to \infty

, toh charging practically kabhi khatam nahi hoti; kisi bhi finite time mein voltage barely badhti hai. Bahut bada capacitor kaafi time tak ek short (ek simple wire jisme

V_{C} \approx 0

) ki tarah lagta hai.

If you connect an already-charged capacitor to a battery of the same voltage

E

, what current flows?

Har waqt zero. Net driving voltage

E - V_{C} = 0

shuruat se, toh

I = 0

— circuit apni steady state mein hi paida hota hai aur kuch nahi badalta.

t \to \infty

during charging, what are

I

V_{C}

, and

V_{R}

I \to 0

V_{C} \to E

, aur

V_{R} = I R \to 0

. Capacitor poora emf hold karta hai aur open switch ki tarah act karta hai; koi current nahi matlab koi resistor drop nahi.

What if you discharge a capacitor through zero resistance?

τ \to 0

ek ideal instantaneous dump aur infinite momentary current deta hai. Physically yeh ek spark/short hai; real resistance aur inductance ise finite rakhte hain.

If two identical resistors are in series in the charging loop, how does

τ

change?

Effective resistance double hokar

2 R

ho jaati hai, toh

τ = 2 R C

— charging mein double waqt lagta hai. Sirf loop mein net series resistance matter karti hai, Ohm's Law and Resistance ke hisaab se.

Recall Ek-line self-test

Charge tak 63% badhta hai, tak 37% girta hai, aur $5 τ$ ke baad practically khatam — in teeno mein se kaun sa tumne upar sabse zyada galat kiya? Answer ::: Zyaatar log "63% tak pahunchna vs 37% tak girna" par trip karte hain — yeh complementary hain, same number nahi.

Connections

Parent: RC circuits & τ = RC
Capacitors and Capacitance · Ohm's Law and Resistance · Kirchhoff's Voltage Law
Exponential Decay and Differential Equations · Newton's Law of Cooling
Energy Stored in a Capacitor · LR Circuits