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.

- 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

\(\Delta V_L < 0\)

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

- When voltmeter is connected in parallel across each element, it gives reading of potential difference across each element.
- Consider a circuit in which voltmeter V
_{1,}V_{2 }and V_{3}are connected.

- Voltmeter V
_{1}measures potential difference across inductor (L). - Voltmeter V
_{2}measures potential difference across resistor (R) . - Voltmeter V
_{3}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 \(\left [\dfrac {dI}{dt} = 0 \right]\).

A \(V_1=0\)

B \(V_2=IR\)

C \(V_3=\mathcal{E}\)

D \(V_1=-L+\mathcal{E}\)

- 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=y
_{0}. - 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 = y_{0} and at t = t, y = y

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

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

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

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

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

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

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

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

C \(x=x_0\)

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

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

A 4 m sec

B 3 m sec

C 0.5 m sec

D 2 m sec

- 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 S
_{2}switch is thrown to point "a" and S_{1}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.

A 6 m sec

B 2 m sec

C 8 m sec

D 4 m sec

- Consider a RL circuit as show in figure.

- A switch S
_{2}is at position "a" for a long time due to allow the current to reach its equilibrium value \(\dfrac {\mathcal{E}}{R}\). - Now, switch S
_{2}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, \(I = I_0 \; e^{-t/_\tau}\)

A 2.21 A

B 3.45 A

C 3.67 A

D 1.45 A