**Linked comprehensive Type**
The circuit is shown below.

Currently the switch is in position A and steady current is flowing to the circuit. Capacitor is uncharged. At t=0, switch is instanously moved to position B.

**Question 1** Find the current in the LC circuit for t≥ 0

Here $\omega = \frac {1} {\sqrt LC}$

Solution
**Question 2** Find the charge q on the lower capacitor plate for t≥ 0

Solution
**Question 3** Which of the following is true of the circuit?

a) The electric oscillations of time period $2\pi \sqrt{LC}$ are produced in the LC circuit

b) The total energy remains conserved and is equal to $\frac {1}{2}L \frac {V_0^2}{R}$

c) The maximum current in the circuit is V

_{0}/R

d) None of these

Solution
$T=2\pi \omega =2 \pi \sqrt {LC}$

Total energy=Initial energy in the inductor

$E=\frac {1}{2} LI^2 =\frac {1}{2} L \frac {V_0^2}{R}$

**Question 4**
An LR circuit is connect to Voltage source V

_{0}. Which of the following is true at time equal to time constant

a) The rate at which energy is being stored in Inductor is given by $\frac {.63 V_0^2}{R}$

b) Power loss in the resistor is given by $\frac {.40 V_0^2}{R}$

c) The rate of change of current is $\frac {.37 V_0^2}{R}$

d) Current is 63% of the maximum value

Solution
**Question 5**
A circuit is shown below in Figure

It is given

L=R

^{2}C

Which of the following is true

a) The total current will instantly rise to its final value V

_{0}/R when switch is closed

b) The current will zero in LR branch and have value V

_{0}/R in the LC circuit path at t=0

c) The current will rise with time in the LR circuit and decrease with time in LC circuit

d) Total current in the circuit will always be V

_{0}/R for t>0

Solution
Let I

_{L} denote the current in the RL branch

And I

_{c} denote the current in RC Branch

Then Total current through battery

I=I

_{L} + I

_{C}
Now since the battery is across two branches in parallel the current through the RL branch is unaffected by the presence of the RC branch ,

Now it is given

L=R

^{2}C

Or

$\frac {L}{R}=RC$

So

$I=\frac {V_0}{R}$

So I is independent of time and equal to V

_{0}/R for t ≥0

Also (1) and (2)

The current will zero in LR branch and have value V

_{0}/R in the LC circuit path at t=0

The current will rise with time in the LR circuit and decrease with time in LC circuit

**Linked comprehensive Type**
A sinusoidal voltage of frequency f=60Hz and peak value 150V is applied to a series LR circuit

Where L=40mH and R=20ohm

**Question 6 Which** one of the following is correct

**Question 7** At t=T/6, which one of the following is true

a) The instantaneous current is 2.344 A

b) The instantaneous voltage across inductor is 83.29 V

c) The instantaneous voltage across resistor is 46.88 V

d) The total voltage applied is 130 V

**Question 8** Which one of the following is true about the circuit?

a) Average power into the circuit is 360 W

b) I

_{rms} is 4.24 A

c) V

_{rms } is 106.07 V

d) None of these

Solutions for Questions 6-8
a) Now from the formula’s

$T=\frac {1} {f}=\frac {1}{60} sec$

$\omega =2\pi f= 377rad/sec $

$X_L=\omega L=15.08 \Omega $

$Z=\sqrt {R^2 + X_L^2} =25.05 \Omega $

$\phi = tan ^-1 \frac {X_L){R}=37^0 $

Now max current

$I=\frac {V}{Z}=150/25.05=6A$

Max voltage across Resistor is

V_{R}=IR=120V

Max Voltage across Inductor is

V_{L}=IX_{L} =90.5V

b) At t=T/6

We have

$\omega t=\frac {\pi }{3}=60^0$

Instantaneous current is given by

$i=I sin (\omega t - \phi) =6 sin 23=2.344 A$

Instantaneous voltage across Inductor

$v=V_L cos (\omega t - \phi)=90.5 cos 23=83.29 V$

Instantaneous voltage across Resistor

v_{R} = iR==46.88

Total voltage =v_{L} + v_{R} =130 V

c)

From the formula’s

$I_rms= \frac {I}{\sqrt 2} =\frac {6}{\sqrt 2} =4.24 A$

$V_rms=\frac {V}{\sqrt 2} = 106.07 V$

Average power= (I_{rms} )^{2} R=360 W

**Question 9**:

A coil of inductance .50H and resistance 100 ohm is connected to 240 V,50Hz AC supply

Which of the following is true

a) I

_{rms} in the coil is 1.28 A

b) The time lag between voltage maximum and current maximum is 3.2X10

^{-3} s

c) Maximum Voltage is 339.36 V

d) None of these

Solution
Given

V

_{rms}=240 V ,L=.50H and R=100ohm

Now

Now for time lag,

Voltage of an ac source varies as

$V=V_0 cos \omega t$

And similarly current varies as

$I=I_0 cos (\omega t - \phi )$

Where

$tan \phi =\frac {\omega L}{R}$

Now for t=0,Voltage is maximum

I will be maximum when ωt-Φ=0 i.e when t=Φ/ω

So that time lag between voltage maximum and current maximum is

$\Delta t=\frac{ \phi }{ \omega }$

Now let us calculate Φ

$tan \phi =\frac {\omega L}{R} =\frac {2 \pi (50)(.5)}{100}=1.57$

Or

Φ=57.5

^{0}=57.5Xπ/180 rad

So time lag

$\Delta t=\frac{ \phi }{ \omega } =3.2 x 10^-3 s$

**Question 10** Consider a series RC circuit for which R = 2.0 × 10

^{6}, C = 6.0 µF and E = 20 V. Find

(i) The time constant of the circuit,

(ii) The maximum charge on the capacitor after a switch in the circuit is closed, and

(iii)

The current in the circuit at the instant just after the switch in the circuit is closed

Class 12 Maths
Class 12 Physics