First slide

Power in a.c. circuit

Question

A light bulb and an inductor coil are connected to an ac source through a key as shown in the figure below. The key is closed and after sometime an iron rod is inserted into the interior of the inductor. The glow of the light bulb

Moderate

A

increases
B

decreases
C

remains unchanged
D

will fluctuate

Solution

As an iron rod is inserted, inductance of inductor will increase. Hence impedance $z = \sqrt{R^{2} + (ω L)^{2}}$ will increase, so from $i = \frac{v}{z}$ current will decrease and from $P = \frac{V_{0}^{2} R}{{2 z}^{2}}$ power of bulb will decrease.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App