Informative line

### R L Circuit

Learn how to determine the time constant of an RL circuit and growth of current in an RL circuit and Self Inductance. Practice to find voltmeter in the circuit to read potential difference.

# Self Inductance (L)

• A current in a closed conducting loop produces a magnetic field around it.
• This magnetic field has its own flux through area bounded by the loop.
• Thus if the current changes with time, the flux also changes.

$$\phi _m\propto \,I$$

$$\phi _m=LI$$

where $$L=$$ Self Inductance

• As the current increases in the loop, the magnetic field increases and thus the flux also increases.

• The induced emf due to change of current in the coil is given as

$$\mathcal{E}=-\dfrac{d\phi}{dt}$$

$$\mathcal{E}=-\dfrac{d}{dt}(LI)$$

$$\mathcal{E}=-L\dfrac{d\,I}{dt}$$

Units of inductance

Units of $$L-Wb/A$$ or Henery $$(H)$$

$$\Delta V_L = -L\dfrac {dI}{dt}$$

• This is the value of potential difference across an inductor, measured along direction of current. ( By sign convention )

Case 1.     Potential difference measured along the direction of current and current is increasing.

Since current is increasing, the potential decreases.

$$\dfrac {dI}{dt}> 0$$

Since, $$\Delta V_L = -L\dfrac {dI}{dt}$$

So, $$\Delta V_L < 0$$     $$\left [since, \dfrac {dI}{dt} > 0\right ]$$

$$\Delta V_L = -L\dfrac {dI}{dt}$$

Note → $$\Delta V_L$$ = – ( positive quantity )

$$\Delta V_L$$ =  (negative quantity )

NOTE: This case is similar to the case when an object freely falls under gravity(downwards),its speed increases and vice versa.

Case 2.      When potential difference is measured along the direction of current and current is decreasing.

• Since, current is decreasing.

$$\dfrac {dI}{dt}< 0$$

• So, potential difference across inductor

$$\Delta V_L = -L\dfrac {dI}{dt}$$

$$\Delta V_L$$ = – ( Negative quantity )

$$\Delta V_L$$ = Positive quantity

$$\therefore$$ $$\Delta V_L > 0$$

Case 3.     Potential difference is measured opposite to direction of current and current is increasing.

• Since, current is increasing .

$$\dfrac {dI}{dt}> 0$$

• Potential difference across inductor

$$\Delta V_L = L\dfrac {dI}{dt}$$  [ Moving opposite to direction of current ]

$$\Delta V_L$$ = $$L\times$$ ( Positive quantity )

$$\Delta V_L$$ = Positive

$$\Delta V_L > 0$$

Case 4.     Potential difference measured opposite to direction of current and current is decreasing.

• Since, current is decreasing.

$$\dfrac {dI}{dt}< 0$$

• Potential difference across inductor

$$\Delta V_L = L\dfrac {dI}{dt}$$

$$\Delta V_L$$ = $$L\times$$ (Negative quantity )

$$\Delta V_L$$ = Negative

$$\Delta V_L < 0$$

#### An inductor coil has its ends labeled a and b. The potential at a is higher than at b. Choose the correct option regarding this situation.

A The current is increasing and directed from a to b

B The current is decreasing and directed from a to b

C The current is constant and directed from a to b

D The current is constant and directed from b to a

×

Potential decreases in the direction of increasing current and current always flow form high to low potential.

Hence option  (A)  is correct.

### An inductor coil has its ends labeled a and b. The potential at a is higher than at b. Choose the correct option regarding this situation.

A

The current is increasing and directed from a to b

.

B

The current is decreasing and directed from a to b

C

The current is constant and directed from a to b

D

The current is constant and directed from b to a

Option A is Correct

# Voltmeter in the Circuit to Read Potential Difference

• When voltmeter is connected in parallel across each element, it gives reading of potential difference across each element.
• Consider a circuit in which voltmeter V1, Vand V3 are connected.

• Voltmeter V1 measures potential difference across inductor (L).
• Voltmeter V2 measures potential difference across resistor (R) .
• Voltmeter V3 measures potential difference across battery ($$\mathcal{E}$$).

Note :-  When battery is connected across inductor L, then potential difference across inductor $$\Delta V_L = 0$$, since the steady current flows through inductor $$\left [\dfrac {dI}{dt} = 0 \right]$$.

#### Choose the incorrect reading of voltmeter in the following circuit at steady state.

A $$V_1=0$$

B $$V_2=IR$$

C $$V_3=\mathcal{E}$$

D $$V_1=-L+\mathcal{E}$$

×

At steady state, the current will be steady current and will not change with time.

The potential drop across resistor $$V_2=IR$$

The potential across battery $$V_3=\mathcal{E}$$

Due to battery, the current is steady and unchanging.

So, $$\dfrac {dI}{dt} = 0$$

Potential difference across inductor,

$$\Delta V_L = V_1 = -L\dfrac{dI}{dt}$$

$$\Delta V_L = V_1 = 0$$

Thus, option (D) is incorrect.

### Choose the incorrect reading of voltmeter in the following circuit at steady state.

A

$$V_1=0$$

.

B

$$V_2=IR$$

C

$$V_3=\mathcal{E}$$

D

$$V_1=-L+\mathcal{E}$$

Option D is Correct

# Growth of Current in an RL Circuit

• Consider a circuit consisting of a battery with negligible resistance, an inductor of inductance L and a resistor of resistance R, as shown in figure.

• Suppose S2 switch is thrown to point "a" and S1 is closed at time t = 0, the current starts increasing in the circuit.

• By applying Kirchoff's law ,

$$\mathcal{E}- IR -L \dfrac {dI}{dt}= 0$$

Rearranging the equation,

$$\mathcal{E} - IR = L\dfrac {dI}{dt}$$

$$\dfrac {dI}{\mathcal{E}- IR}\;\;=\;\;\dfrac {dt}{L}$$

• Integrating both sides and applying limits,

$$\int\limits^{I}_0\;\dfrac {dI}{\mathcal{E}-IR}\;=\; \int\limits^{t}_0 \dfrac {dt}{L}$$

Let $$x = \mathcal{E}- IR$$,    then $$dx=-RdI$$

$$\dfrac {-1}{R}\;\;\int\limits^{\mathcal{E}-IR}_\mathcal{E}\dfrac {dx}{x}\;\;=\;\;\dfrac {1}{L}\int\limits^{t}_0dt$$

$$\Rightarrow \;\;\dfrac {-1}{R}[lnx]^{\mathcal{E}-IR}_\mathcal{E} = \dfrac {t}{L}$$

$$\Rightarrow \;\;ln \dfrac {\mathcal{E}-IR}{\mathcal{E}}= \dfrac {-Rt}{L}$$

$$\Rightarrow \;\; \dfrac {\mathcal{E}-IR}{\mathcal{E}} = e^{-Rt/L}$$

$$\Rightarrow \mathcal{E} - IR = \mathcal{E} \;e^{-t/(L/R)}$$

$$\Rightarrow\;\;I = \dfrac {\mathcal{E} }{R}[1-e^{-t/(L/R)}]$$

• This expression shows that current does not increase instantly to its final equilibrium value as switch is closed. It takes infinite time to reach steady state.

• By time constant $$\tau = \dfrac {L}{R}$$

$$I = \dfrac {\mathcal{E}}{R}[1-e^{-t/\tau}]$$

• When time constant  $$\tau = t$$

$$I = \dfrac {\mathcal{E}}{R}[1-e^{-t/t}]$$

$$I = \dfrac {\mathcal{E}}{R}[1-e^{-1}]$$

$$I = \dfrac {\mathcal{E}}{R}[1-.37]$$

$$I = \dfrac {\mathcal{E}}{R} × 0.632$$

$$I=63.2\ \%$$ of its final value.

#### During growth of current in an RL circuit, at what time current will be 63.2 % of its final value for the following circuit.

A 6 m sec

B 2 m sec

C 8 m sec

D 4 m sec

×

For current to be 63.2% of the final value,

Time constant of RL circuit = t

Time $$t=\tau = \dfrac {L}{R}$$

where ,

L = Inductance

R = Resistance

$$t = \dfrac {4 × 10 ^{-3}}{2}$$

$$t=2\ m\ sec$$

### During growth of current in an RL circuit, at what time current will be 63.2 % of its final value for the following circuit.

A

6 m sec

.

B

2 m sec

C

8 m sec

D

4 m sec

Option B is Correct

# Decay of Current in an RL Circuit

•  Consider a RL circuit as show in figure.

• A switch S2 is at position "a" for a long time due to allow the current to reach its equilibrium value $$\dfrac {\mathcal{E}}{R}$$.
• Now, switch S2 is thrown from a to b.
• Applying Kirchoff's law,

$$IR + L\dfrac {dI}{dt} = 0$$

$$L\dfrac {dI}{dt} = -IR$$

$$\dfrac {dI}{I} = \dfrac {-R}{L}\;\;dt$$

• Integrating both sides

$$\Rightarrow \;\;\int\limits^{I}_{\mathcal{E}/_R} \;\dfrac {dI}{I}\;\;=\;\;\dfrac {-R}{L}\;\int \limits^t_0 dt$$

$$\Rightarrow\;\;[lnI]^{I}_{\mathcal{E}/_R}\;\;=\;\;\dfrac {-R}{L}t$$

$$\Rightarrow\;\;ln \left(\dfrac {I}{\mathcal{E}/_R}\right) = \dfrac {-R}{L}t$$

$$\Rightarrow \dfrac {I}{\mathcal{E}/_R}=e^{-Rt/L}$$

$$\Rightarrow I = \dfrac {\mathcal{E}}{R}\;\;e^{-t/(L/_R)}$$

$$\Rightarrow I = I_0\;\;e^{-t/(L/_R)}$$

• We know ,

Time constant $$\tau = \dfrac {L}{R}$$

So, $$I = I_0 \; e^{-t/_\tau}$$

#### During the discharging of the following circuit, determine the current in the circuit at the time $$t= \tau$$ = time constant.

A 2.21 A

B 3.45 A

C 3.67 A

D 1.45 A

×

The instantaneous current during discharging is given as

$$I = I_0 \; e^{-t/_\tau}$$

where, $$I_0 = \dfrac {\mathcal{E}}{R}$$

Given :   $$L=4\ mH$$,    $$R=2\ \Omega$$,    $$\mathcal{E}=12\ V$$

$$I = \dfrac {12}{2}\;e^{-\tau/\tau}$$

$$I=6\,e^{-1}$$

$$I=6\times0.369$$

$$I=2.21\ A$$

### During the discharging of the following circuit, determine the current in the circuit at the time $$t= \tau$$ = time constant.

A

2.21 A

.

B

3.45 A

C

3.67 A

D

1.45 A

Option A is Correct

# Calculation of "y" as a Function of Time Using Integration

• Given an equation  $$ay + \dfrac {bdy}{dt} = 0$$
• To calculate "y" as a function of time using integration, the limits will be at t=0, y=y0.
• Separation of variable is to be done.

$$ay = \dfrac {-bdy}{dt}$$

$$\Rightarrow \dfrac{-a}{b}\;y = \dfrac {dy}{dt}$$

$$\Rightarrow \dfrac{-a}{b}\;dt = \dfrac {dy}{y}$$

Integrating both sides by applying limits,

at   t = 0,  y = y0     and     at  t = t,   y = y

$$\Rightarrow \dfrac{-a}{b} \int\limits^{t}_0 dt\;\;=\;\;\int\limits^{y}_{y_0} \;\; \dfrac{dy}{y}$$

$$\Rightarrow \dfrac{-a}{b}t\;\;=\;[ln\,y ]^{y}_{y_0}$$

$$\Rightarrow \dfrac{-a}{b}t\;\;=\;\;ln\,y\;-\;ln\,y_0$$

$$\Rightarrow \dfrac{-a}{b}t\;\;=\;\;ln \;\left (\dfrac{y}{y_0}\right)$$

$$\Rightarrow \dfrac{y}{y_0}= e^{-{\left(\dfrac {t}{b/a}\right)}}$$

$$\Rightarrow y = y_0\;\;\;e^{-{ \left (\dfrac{t}{b/_a}\right) }}$$

#### Determine the value of $$x$$ as a function of time for the following expression  $$2 x + 4 \dfrac{dx}{dt} = 0$$ If at   $$t=0$$,   $$x = x_0$$.

A $$x=x_0\;e^{-4t}$$

B $$x=x_0\;e^{-2t}$$

C $$x=x_0$$

D $$x=x_0\;e^{-{t/_2}}$$

×

Separation of variable ,

$$2x = \dfrac{-4\;dx}{dt}$$

$$x = \dfrac {-2\; dx}{dt}$$

$$\dfrac{-dt}{2}\;\;=\;\;\dfrac{dx}{x}$$

Integrating both sides and applying limits,

at   $$t=0$$,    $$x$$ =$$x_0$$

at $$t=t$$,    $$x$$ =  $$x$$

$$\int\limits^{t}_0 \dfrac{-dt}{2}\;\;=\;\;\int\limits^{x}_{x_0} \;\; \dfrac{dx}{x}$$

$$\Rightarrow \dfrac{-t}{2}\;\;=\;[lnx ]^{x}_{x_0}$$

$$\Rightarrow \dfrac{-t}{2}\;\;=\;\;ln \;\dfrac{x}{x_0}$$

$$\Rightarrow \dfrac{x}{x_0}= e^{-t/_2}$$

$$\Rightarrow {x} = {x_0}\; e^{-(t/_2)}$$

### Determine the value of $$x$$ as a function of time for the following expression  $$2 x + 4 \dfrac{dx}{dt} = 0$$ If at   $$t=0$$,   $$x = x_0$$.

A

$$x=x_0\;e^{-4t}$$

.

B

$$x=x_0\;e^{-2t}$$

C

$$x=x_0$$

D

$$x=x_0\;e^{-{t/_2}}$$

Option D is Correct

# Time Constant of an RL Circuit

• Time constant of an RL circuit is defined as the ratio of self inductance (L) of the inductor and resistance(R) of the resistor.
• Time constant is denoted by $$\tau$$.
• Time constant of an RL circuit is given as

$$\tau = \dfrac {L}{R}$$

## S.I. Unit of Time Constant

• S.I. unit of time constant is same as S.I unit of time ,i.e., ''second''.

#### Determine the time constant of an RL circuit, as shown in figure.

A 4 m sec

B 3 m sec

C 0.5 m sec

D 2 m sec

×

Time constant $$\tau$$ of an RL circuit is given as

$$\tau = \dfrac {L}{R}$$

where,

L = Inductance

R = Resistance

Given: $$L=3\ mH$$,  $$R=6\ \Omega$$

$$\tau = \dfrac {3 × 10^{-3}}{6} = 0.5\ m\;sec$$

### Determine the time constant of an RL circuit, as shown in figure.

A

4 m sec

.

B

3 m sec

C

0.5 m sec

D

2 m sec

Option C is Correct