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

Impedance Triangle

In the circui...

Question

In the circuit shown below, the key $K$ is closed at $t = 0$ . The current through the battery is

\frac{V R_{1} R_{2}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}

t = 0

and

\frac{V}{R^{2}}

t = \infty

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

\frac{V}{R_{2}}

t = 0

and

\frac{V (R_{1} + R_{2})}{R_{1} R_{2}}

t = \infty

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

\frac{V}{R^{2}}

t = 0

and

\frac{V R_{1} R_{2}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}

t = \infty

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

\frac{V (R_{1} + R_{2})}{R_{1}^{2} R_{2}}

t = 0

and

\frac{V}{R^{2}}

t = \infty

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

Open in App

Solution

The correct option is B $\frac{V}{R_{2}}$ at $t = 0$ and $\frac{V (R_{1} + R_{2})}{R_{1} R_{2}}$ at $t = \infty$
We know that,

At $t = 0$ , inductor behave like an infinite resistance

So, at $t = 0$

$i = \frac{V}{R_{2}}$

Now, at $t = \infty$ , inductor behave like conductive wire

So,

$i = \frac{V (R_{1} + R_{2})}{R_{1} R_{2}}$

Hence, the current is $i = \frac{V}{R_{2}}$ at $t = 0$ and $\frac{V (R_{1} + R_{2})}{R_{1} R_{2}}$ at $t = \infty$

Suggest Corrections

Similar questions

K

t = 0

S_{1}

t = 0

t = (2 R_{2} + R_{1}) C, S_{1}

S_{2}

t = (2 R_{2} + 2 R_{1}) C

R_{2}

t = (3 R_{1} + 2 R_{2}) C

V_{2}

R_{2}

t = 0

t = \infty

t = 0

t = 10^{- 3} l n (2)

6 μ F

160 μ C

s

t = 0

R_{1}

R_{2}

t = 16 μ s

Join BYJU'S Learning Program