4μF capacitor and a resistance 2.5 MΩ are in series with 12 V battery. Find the time after which the potential difference…

Q: 4μF capacitor and a resistance 2.5 MΩ are in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [given in (2)= 0.693]

Sol: (a) Charging current $\large I = \frac{V_0}{R} e^{\frac{-t}{R C}}$

Potential difference across R is

$\large V_R = I R = V_0 e^{\frac{-t}{R C}} $

Potential difference across ‘C’ is

$\large V_C = V_0 – V_R = V_0 – V_0 e^{\frac{-t}{R C}} $

$\large V_C = V_0 (1 – e^{\frac{-t}{R C}}) $

But given VC = 3 VR , we get

$\large V_0 (1 – e^{\frac{-t}{R C}}) = 3 V_0 e^{\frac{-t}{R C}} $

$\large (1 – e^{\frac{-t}{R C}}) = 3 e^{\frac{-t}{R C}} $

$\large 1 = 4 e^{\frac{-t}{R C}} $

$\large e^{\frac{t}{R C}} = 4 $

$\large \frac{t}{R C} = ln 4 $

$\large t = 2 R C ln2 $

t = 2 × 2.5 × 106 × 4 × 10-6 × 0.693

or , t = 13.86 sec.

A parallel – plate capacitor, filled with a dielectric of dielectric constant k, is charged to a potential V0…

Q: A parallel – plate capacitor, filled with a dielectric of dielectric constant k, is charged to a potential V0. It is now disconnected from the cell and the slab is removed. If it now discharges, with time constant τ , through a resistance, then find time after which the potential difference across it will be V0 ?

Sol: When slab is removed, the potential difference across capacitor increases to k V0

$\large C V_0 = k C V_0 e^{\frac{-t}{\tau}}$ ; as q0 = k C V0

$\large \frac{1}{k} = e^{\frac{-t}{\tau}}$

$\large k = e^{\frac{t}{\tau}}$

$\large lnk = \frac{t}{\tau}$

$\large t = \tau lnk $

A coil of resistance 20 Ω and inductance 0.5 henry is switched to dc 200 volt supply. Calculate the rate of increase of current …

Q: A coil of resistance 20 Ω and inductance 0.5 henry is switched to dc 200 volt supply. Calculate the rate of increase of current:
(a) At the instant of closing the switch and
(b) After one time constant
(c) Find the steady state current in the circuit.

Sol: (a) This is the case of growth of current in an L – R circuit. Hence, current at time t is given

$\large I = I_0 (1-e^{\frac{-t}{\tau}})$

Rate of increase of current,

$\large \frac{dI}{dt} = \frac{I_0}{\tau} e^{\frac{-t}{\tau}}$

At , t = 0 ;

$\large \frac{dI}{dt} = \frac{I_0}{\tau} = \frac{E/R}{L/R}$

$\large \frac{dI}{dt} = \frac{E}{L} = \frac{200}{0.5}$

= 400 A/s

(b) At , t = τ ;

$\large \frac{dI}{dt} = \frac{I_0}{\tau} e^{-1}$

$\large \frac{dI}{dt} = 400 e^{-1} = 400 \times 0.37 $

= 148 A/s

(c) The steady state current in the circuit is

$\large I_0 = \frac{E}{R} = \frac{200}{20} $

= 10 A

Calculate the back e.m.f of a 10 H, 200 Ω coil 100 ms after a 100 V d.c supply is connected to it.

Q: Calculate the back e.m.f of a 10 H, 200 Ω coil 100 ms after a 100 V d.c supply is connected to it.

Sol: The value of current at 100 ms after the switch is closed is

$\large I = I_0 (1-e^{\frac{-t}{\tau}})$

Here , $I_0 = \frac{E}{R} = \frac{100}{200} $

= 0.5 A

$\large \tau = \frac{L}{R} = \frac{10}{200} $

= 0.05 sec ; t = 100 ms = 0.1 sec

$\large I = 0.5(1-e^{\frac{-0.1}{0.05}})$

$\large I = 0.5(1-e^{-2})$ = 0.4325 A

$\large E = I R + L \frac{dI}{dt}$

$\large 100 = 0.4325 \times 200 + L \frac{dI}{dt}$

Back e.m.f $\large = L \frac{dI}{dt} = 100 – 0.4325 \times 200 $

= 13.5 volt

A coil having resistance 15 Ω and inductance 10 H is connected across a 90 Volt dc supply. Determine the value of current after 2 sec…

Q: A coil having resistance 15 Ω and inductance 10 H is connected across a 90 Volt dc supply. Determine the value of current after 2 sec. What is the energy stored in the magnetic field at that instant.

Sol: Give that; R = 15 Ω, L = 10 H, E = 90 Volt

Peak value of current , $\large I_0 = \frac{E}{R} = \frac{90}{15}$

= 6 A

Also , $\large \tau = \frac{L}{R} = \frac{10}{15} $

= 0.67 sec

$\large I = I_0 (1-e^{\frac{-Rt}{L}})$

After 2 sec ,

$\large I = 6 (1-e^{\frac{-2}{0.67}})$

= 6 (1-0.05) = 5.7 A

Energy stored in a magnetic field is

$\large U = \frac{1}{2}L I^2 $

$\large U = \frac{1}{2}\times 10(5.7)^2 $

= 162.45 J

A cell of 1.5 V is connected across an inductor of 2 mH in series with a 2 Ω resistor. What is the rate of growth of current…

Q: A cell of 1.5 V is connected across an inductor of 2 mH in series with a 2 Ω resistor. What is the rate of growth of current immediately after the cell is switched on.

Sol: $\large E = L\frac{dI}{dt} + I R $

Therefore , $\large \frac{dI}{dt} = \frac{E-IR}{L}$

E = 1.5 volt , R = 2Ω , L = 2 mH = 2 × 10-3 H when the cell is switched on, I = 0

$\large \frac{dI}{dt} = \frac{E}{L}$

$\large \frac{dI}{dt} = \frac{1.5}{2 \times 10^{-3}}$

= 750 A/s

An indicator of 3 H is connected to a battery of emf 6 V through a resistance of 100 Ω. Calculate the time constant…

Q: An indicator of 3 H is connected to a battery of emf 6 V through a resistance of 100 Ω. Calculate the time constant. What will be the maximum value of current in the circuit ?

Sol: Give that L = 3H , E = 6V , R = 100 Ω

Time Constant $\large \tau = \frac{L}{R}$

$\large \tau = \frac{3}{100}$

= 0.03 sec

Maximum Current $\large I_0 = \frac{E}{R} $

$\large I_0 = \frac{6}{100} $

= 0.06 volt