Informative line

### Rc Circuit

Learn potential difference across capacitor and RC circuit with problems practice. Practice to calculate voltage across capacitor in RC circuit and time constant of RC circuit, and solving of integration of similar function as charging and discharging of capacitor.

# Potential Difference Across Capacitor

• Consider a capacitor as shown in figure.

• Potential difference across terminal A and B is given as.

$$\Delta V = V_A - V_B$$

•  Hence, it is clear that potential difference across a capacitor depends upon polarity.
• Case 1 :    The direction of current flow is from A to B, as shown in figure.

$$\Delta V = V_A - V_B$$

• Case 2 :   The direction of current flow is from B to A,  as shown in figure.

$$\Delta V = V_A - V_B$$

Conclusion:

The potential difference across the capacitor only depends upon the polarity and is independent of direction of current flowing through it.

#### Which of the point is at higher potential?

A Point A

B Point B

C Both  A and  B are at same potential

D

×

Since, A is connected to positive terminal of capacitor. So, point A is at higher potential.

Hence, option (A) is correct.

### Which of the point is at higher potential?

A

Point A

.

B

Point B

C

Both  A and  B are at same potential

D

Option A is Correct

# Voltmeter Reading Across Capacitor and Resistor

• Consider a circuit consisting of a resistor, a capacitor and a battery.

• The current $$I$$ is flowing in the circuit and the charge on the plates of capacitor is q.
• Let two voltmeters are connected across resistor and a capacitor, as shown in figure.

• The reading of the voltmeter gives the potential difference across the elements and it is always connected in parallel with elements.
• The reading of voltmeter $$V_1$$

$$V_1 = I\;R$$

• The reading of voltmeter $$V_2$$

$$V_2 = \dfrac{q}{C}$$

#### Choose the incorrect option regarding reading of voltmeter.

A Reading of $$V_1$$ = Reading of $$V_2$$

B Reading of $$V_1$$$$=9\,V$$

C Reading of $$V_2$$$$= 2\,V$$

D Reading of  $$V_1= 5 V$$

×

Reading of voltmeter  $$V_1$$

Voltage across resistor is given as

$$V_1 = I R$$

where $$I =$$    current ,

$$R =$$ Resistance

$$V_1$$ $$= 3 × 3 = 9\,V$$

Reading of voltmeter  $$V_2$$

Voltage across capacitor is given as

$$V_2 = \dfrac {q}{C}$$

[where q = charge on capacitor , C= capacitance]

or, $$V_2 = \dfrac {1\mu C}{1\mu F}$$

or, $$V_2 =1\,V$$

### Choose the incorrect option regarding reading of voltmeter.

A

Reading of $$V_1$$ = Reading of $$V_2$$

.

B

Reading of $$V_1$$$$=9\,V$$

C

Reading of $$V_2$$$$= 2\,V$$

D

Reading of  $$V_1= 5 V$$

Option B is Correct

# Charging of Capacitor in an RC Circuit

• A circuit consisting of a resistor and a capacitor is known as an RC circuit.

## Charging of Capacitor

• Consider a circuit consisting of resistor (R) and capacitor (C), connected with a battery ($$\mathcal{E}$$), as shown in figure.

• Initially the capacitor of the circuit is uncharged as the switch is open.
• Now switch is thrown to position 'a' at t = 0.
• When switch is closed, the charge starts to flow in the circuit, setting up the current and the capacitor begins to charge.
• As capacitor is charging, potential difference across the capacitor increases.
• After some time when capacitor is fully charged,  the current in the circuit becomes zero as the potential difference across the capacitor becomes same as that of battery e.m.f.
• By applying Kirchhoff's law in the circuit.,

$$\mathcal{E} -\dfrac{q}{C} - IR = 0$$              ..........(1)

Note -   Here, $$q$$ and $$I$$ are instantaneous values.

Case 1 :

At  time t=0, when switch is just closed, the capacitor is uncharged. So, charge on capacitor is zero.

$$q_i = 0$$

From (1)

So, $$I_i=\dfrac{\mathcal{E}}{R}$$

• At time t=0 , the potential difference from the battery terminal appears entirely across the resistor.

Case 2:

When capacitor is fully charged, the current in the circuit becomes zero.

$$I =0$$

From equation (1)

$$Q = C\mathcal{E}$$ (Maximum value)

• To analyze time dependence of charge and current

put $$I = \dfrac{dq}{dt}$$ in equation (1)

Then      $$\dfrac{dq}{dt} = \dfrac{\mathcal{E}}{R} - \dfrac{q}{RC}$$

or,      $$\dfrac{dq}{dt} = \dfrac{ C\mathcal{E}}{RC} - \dfrac{q}{RC}$$

or,      $$\dfrac{dq}{dt} = -\left(\dfrac{q-C \mathcal{E}}{RC} \right)$$

or,      $$\dfrac{dq}{q-C\mathcal{E}} = \dfrac{-1}{RC} dt$$

•  Integrating both sides

$$\int\limits^{q}_o \dfrac{dq}{(q-C\mathcal{E})} = \dfrac{-1}{RC}\int\limits^{t}_o dt$$

or,    $$ln \left( \dfrac{q-C\mathcal{E}}{-C\mathcal{E}}\right) = \dfrac{-t}{RC}$$

or,      $$\dfrac{q- C \mathcal{E}}{-C\mathcal{E}} = e^{-t/RC}$$

or,      $$q(t) = C\mathcal{E}(1-e^{-t/RC})$$

$$q(t) = Q_{max} (1-e^{-t/RC})$$

•    Time constant for an  RC circuit is given as

$$\tau =RC$$

•  So, the charge on capacitor at any instant is

.           $$q(t) = Q_{max} (1- e ^{-t/\tau})$$

#### An uncharged capacitor $$C = 5\,\mu F$$ and a resistor $$R = 8 ×10^5 \,\Omega$$, are connected in series with a battery of e.m.f. $$\mathcal{E} = 12\,V$$. Calculate the instantaneous charge on the capacitor when switch is thrown to position 'a' at any instant time t.

A $$60\ e^{-0.25t} \,\mu C$$

B $$60(1-e^{-0.25t} )\,\mu C$$

C $$60 \,\mu C$$

D $$60 -e^{-0.25t} \,\mu C$$

×

Maximum charge that can be acquired by capacitor, is given as

$$Q_{max} = C \mathcal{E}$$

or,    $$Q_{max} = 5× 10^{-6} ×12$$

or,     $$Q_{max} = 60 \,\mu C$$

Charge on capacitor at any instant of time t is given as

$$q(t) = Q_{max} (1- e ^{-t/\tau})$$

where $$\tau = RC$$

Time constant $$(\tau)$$

$$\tau =RC$$

$$\tau = 8 × 10^5 × 5 ×10^{-6}$$

$$\tau$$ $$= 4\, sec$$

$$q(t)= 60 ×10^{-6} (1-e^{-t/4})$$

$$q(t)= 60 (1-e^{-0.25t}) \,\mu C$$

### An uncharged capacitor $$C = 5\,\mu F$$ and a resistor $$R = 8 ×10^5 \,\Omega$$, are connected in series with a battery of e.m.f. $$\mathcal{E} = 12\,V$$. Calculate the instantaneous charge on the capacitor when switch is thrown to position 'a' at any instant time t.

A

$$60\ e^{-0.25t} \,\mu C$$

.

B

$$60(1-e^{-0.25t} )\,\mu C$$

C

$$60 \,\mu C$$

D

$$60 -e^{-0.25t} \,\mu C$$

Option B is Correct

# Discharging of a Capacitor in an RC Circuit

• Consider a circuit consisting of a capacitor and a resistor connected, as shown in figure.
• Initially the switch is at 'a' and the capacitor is charged.
• The current reduces to zero when capacitor is fully charged $$(I=0)$$.
• Then potential difference across capacitor is givens as

$$\Delta V = \dfrac{Q}{C}$$

• Now switch is moved from position a to b at t = 0.
• The capacitor begins to discharge through the resistor and the current starts flowing.

• Let at some time t, current is $$I$$ and charge is $$q$$ on the capacitor.
• Applying Kirchhoff's Voltage Law in loop through which capacitor is discharging

$$\dfrac{-q}{C} +IR = 0$$   ......(1)

• Also current in terms of charge is given as

$$I = \dfrac{-dq}{dt}$$   .........(2)

From  (1) and (2)

$$-R \dfrac{dq}{dt} = \dfrac{q}{C}$$

$$\Rightarrow\dfrac{dq}{q} =- \dfrac{1}{RC} dt$$

• Since charge on capacitor varies from maximum charge Q to q at time t

$$\int\limits^q_{Q} \dfrac{dq}{q}= -\dfrac{1}{RC} \int \limits^t _{0} dt$$

$$\Rightarrow \left [ ln(q)\right ]^q_Q = -\dfrac{1}{RC} \int \limits^t_{0} dt$$

$$\Rightarrow \ ln(q) - ln(Q) = \dfrac{-t}{RC}$$

$$\Rightarrow \ ln \left(\dfrac{q}{Q}\right) = \dfrac{-t}{RC}$$

$$\Rightarrow \ q(t) = Q\,e^{-t/RC}$$

$$\Rightarrow \ q(t) = Q\,e^{-t/\tau}$$      where $$\tau= RC$$

#### Calculate the value of charge at any instant of time 't' when switch is thrown at point b in the  given circuit. Assume that the capacitor is fully charged and $$C= 2$$ $$\mu F$$, $$R =$$$$2 × 10^6 \,\Omega$$ and $$\mathcal{E} = 20\,V$$.

A $$40$$ $$e ^{-t/4} \,\mu C$$

B  $$40\,\mu C$$

C $$e ^{-t} \,\mu C$$

D  $$40e ^{-t} \,\mu C$$

×

Time constant of an RC circuit

$$\tau =RC$$

or, $$\tau = 2 ×10^6×2 × 10^{-6}$$

$$=4\, sec$$

Maximum charge that can be acquired by capacitor is given as

$$Q = C \mathcal{E}$$

or,    $$Q = 2 × 10^{ -6 }× 20$$

or,   $$Q =$$$$40\, \mu C$$

Charge at any instant of time in capacitor is given as

$$q(t) = Q_{max}\, e ^{-t/ \tau}$$

$$\Rightarrow$$$$q(t) = 40 × 10^{-6} e^{-t/4}$$

$$\Rightarrow$$ $$q(t) = 40e^{-t/4}$$ $$\mu C$$

### Calculate the value of charge at any instant of time 't' when switch is thrown at point b in the  given circuit. Assume that the capacitor is fully charged and $$C= 2$$ $$\mu F$$, $$R =$$$$2 × 10^6 \,\Omega$$ and $$\mathcal{E} = 20\,V$$.

A

$$40$$ $$e ^{-t/4} \,\mu C$$

.

B

$$40\,\mu C$$

C

$$e ^{-t} \,\mu C$$

D

$$40e ^{-t} \,\mu C$$

Option A is Correct

# Solving of Integration of Similar Function as Charging and Discharging of Capacitor

• Consider a function

$$\dfrac {d y}{dt} = \dfrac{-(y-ab)}{cd}$$

where a,b,c and d are constant.

• To calculate value of y, several steps are to be followed.

Step 1: Separate the terms of y accordingly, i.e., the variable of y on one side and constant on other side.

$$\dfrac {d y}{(y- ab)} = \dfrac{-dt}{cd}$$

Step 2: Integrating both sides

$$\int \dfrac {d y}{(y- ab)} = -\dfrac{1}{cd} \int dt$$

Step 3 : Applying limits to integration on both sides

$$\int\limits^{y}_o \dfrac {d y}{(y- ab)} = - \dfrac{1}{cd} \int\limits^{t}_odt$$

$$ln\; \left (\dfrac {y- ab}{-ab} \right) = \dfrac{-t}{cd}$$

Step 4 : Taking exponential on both sides;

$$\left( \dfrac {y- ab}{-ab}\right) = e^{- \left (\dfrac{t}{cd}\right)}$$

or,  $$(y- ab) = -ab \;e^ {-\dfrac{t}{cd}}$$

or,   $$y = ab-ab \;e^ {-\dfrac{t}{cd}}$$

or, $$y = ab\left(1- \;e^ {-\dfrac{t}{cd}}\right)$$

#### Integrate the following function for value of Q     $$\int \limits ^Q_P \dfrac{1}{y} dy = \int \limits ^x_o Ndx$$ , where N is constant.

A $$e ^{(Nx+lnP)}$$

B $$ln P + Nx$$

C $$ln P + e^{Nx}$$

D $$e^{Nx} - lnP$$

×

$$\int \limits ^Q_P \dfrac{1}{y} dy = \int \limits ^x_o Ndx$$

$$\Rightarrow\left[ ln(y) \right]^Q_p = \left [Nx \right]^x_o$$

$$\Rightarrow ln(Q) - ln(P) = Nx$$

$$\Rightarrow ln Q= Nx +lnP$$

$$\Rightarrow Q = e^{(Nx + lnP)}$$

### Integrate the following function for value of Q     $$\int \limits ^Q_P \dfrac{1}{y} dy = \int \limits ^x_o Ndx$$ , where N is constant.

A

$$e ^{(Nx+lnP)}$$

.

B

$$ln P + Nx$$

C

$$ln P + e^{Nx}$$

D

$$e^{Nx} - lnP$$

Option A is Correct

# Time Constant of an RC Circuit

• Consider a circuit having a resistor R and a capacitor C, as shown in figure.

• Time constant of an RC circuit is denoted by '$$\tau$$' .
• Time constant of an RC circuit is given as

$$\tau = RC$$

where R = Resistance

C = Capacitance

$$\tau =$$ Time constant

#### Calculate the time constant of an RC circuit having a capacitor $$C= 7\,\mu F$$ and a resistor $$R = 3K\,\Omega$$, connected with e.m.f. of  $$\mathcal{E} = 22\,V$$, as shown in figure.

A 3 ms

B 21 ms

C 7 ms

D 20 ms

×

Time constant of an RC circuit is given as

$$\tau = RC$$

where R = Resistance

C = Capacitance

$$\tau =$$ Time constant

Given : $$R = 3\,K \Omega ,\,\,\, C= 7\,\mu F$$

$$\tau = 3 × 10 ^3 × 7 × 10^{-6}$$

$$\tau = 21\, ms$$

### Calculate the time constant of an RC circuit having a capacitor $$C= 7\,\mu F$$ and a resistor $$R = 3K\,\Omega$$, connected with e.m.f. of  $$\mathcal{E} = 22\,V$$, as shown in figure.

A

3 ms

.

B

21 ms

C

7 ms

D

20 ms

Option B is Correct