CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Power

In the circui...

Question

In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterward, suddenly point C is disconnected from point A and connected to point B at time $t = 0$ . Ratio of the voltage across resistance and the inductor at $t = l / r$ will be equal to

A

\frac{e}{1 - e}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

- 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

\frac{1 - e}{e}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 1$
Applying Kirchoff's law of voltage in closed loop,
$- V_{R} - V_{C} = 0$
$\frac{V_{R}}{V_{C}} = - 1$
Thus, the ratio of voltage across resistance and the inductor at $t = \frac{l}{r}$ is equal to $- 1$

Suggest Corrections

thumbs-up

0

Similar questions

Q. In the circuit shown here, the point

C

is kept connected to point

A

till the current flowing through the circuit become constant. Afterward, suddenly point

C

disconnected from point

A

and connected to point

B

t = 0

. Ratio of the voltage across resistance and the inductor at

t = \frac{L}{R}

will be equal to

Q. In the circuit shown in figure,

E_{1} = 3 V, E_{2} = 2 V, E_{3} = 1 V

r_{1} = r_{2} = r_{3} = 1 o h m

(a) Find the potential difference between the points

A

B

and the currents through each branch.
(b) If

r_{2}

is short circuited and the point

A

is connected to point

B

through a resistance

R

, find the currents through

E_{1}, E_{2}, E_{3}

and the resistor

R

.

Q. The circuit shown here is used to compare the e.m.f.'s of the cells

E_{1}

E_{2} (E_{1} > E_{2})

. When the galvanometer is connected to

E_{1}

, the null point is at

C

. When the galvanometer is connected to

E_{2}

, the null point will be

Q. In the circuit shown here

E_{1} = E_{2} = E_{3} = 2 V

R_{1} = R_{2} = 4 Ω

. The current flowing between point A and B through battery

E_{2}

(L = 100 m H)

(R = 100 Ω)

(E = 100 V)

are initially connected in series shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit,

1 m s

after the short circuit is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon