In the given circuit,the readings of voltmete | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The circuit is an $LCR$ series circuit connected to an $AC$ source of $220 \, V$.Given: $V_L = V_1 = 300 \, V$ and $V_C = V_2 = 300 \, V$.Since $V

Vedclass · Accepted Answer

$220 \, V, \, 2.2 \, A$

Vedclass · Answer

(C) The circuit is an $LCR$ series circuit connected to an $AC$ source of $220 \, V$.Given: $V_L = V_1 = 300 \, V$ and $V_C = V_2 = 300 \, V$.Since $V_L = V_C$,the circuit is in resonance.In a series $LCR$ circuit at resonance,the net voltage across the $LC$ combination is zero $(V_{LC} = V_L - V_C = 300 - 300 = 0 \, V)$.Therefore,the entire source voltage appears across the resistor $R$.Thus,the reading of voltmeter $V_3$ is $V_R = 220 \, V$.The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A$.So,the reading of voltmeter $V_3$ is $220 \, V$ and the reading of ammeter $A$ is $2.2 \, A$.

In the given circuit,the readings of voltmeters $V_1$ and $V_2$ are $300 \, V$ each. The readings of the voltmeter $V_3$ and ammeter $A$ are respectively:

Similar Questions

Ribose is an example of

$Assertion (A)$: van der Waals' forces are responsible for chemisorption. $Reason (R)$: High temperature is favourable for chemisorption. The correct answer is

The base current in a transistor circuit changes from $45 \mu A$ to $140 \mu A$. Accordingly,the collector current changes from $0.2 \text{ mA}$ to $4.0 \text{ mA}$. The current gain is

The transformed equation of $3x^2 + 3y^2 + 2xy = 2$,when the coordinate axes are rotated through an angle of $45^{\circ}$,is

$A$ point source of light is placed $4 \, m$ below the surface of water of refractive index $5/3$. The minimum diameter of a disc that should be placed over the source on the surface of water to cut off all light coming out of the water is ...... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module