The reading of the ammeter in the circuit sho | vedclass

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

Question

(A) In the given circuit,the capacitor $(X_C = 5 \ \Omega)$ and the inductor $(X_L = 5 \ \Omega)$ are connected in series with each other,and this com

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) In the given circuit,the capacitor $(X_C = 5 \ \Omega)$ and the inductor $(X_L = 5 \ \Omega)$ are connected in series with each other,and this combination is connected in parallel with the voltmeter. However,the entire series combination of the $LC$ circuit is in series with the resistor $(R = 55 \ \Omega)$ and the $AC$ source $(110 \ V)$.Since $X_L = X_C = 5 \ \Omega$,the net reactance of the $LC$ branch is $X = |X_L - X_C| = |5 - 5| = 0 \ \Omega$.This means the $LC$ branch acts as a short circuit (zero impedance) at resonance.The total impedance of the circuit is simply the resistance $R = 55 \ \Omega$.Using Ohm's law for the $AC$ circuit,the current $I$ is given by $I = \frac{V}{R} = \frac{110 \ V}{55 \ \Omega} = 2 \ A$.Therefore,the ammeter reading is $2 \ A$.

The reading of the ammeter in the circuit shown will be: (in $A$)

Similar Questions

Let $a, b, c \in R$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy$ for all $x, y \in R$,then $\sum_{n=1}^{10} f(n)$ is equal to:

An $AC$ voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased:

Identify the odd one out.

Which of the following reactions defines $\Delta H_f^o$?

Let $\alpha \neq 1$ be a real root of the equation $x^3-ax^2+ax-1=0$,where $a \neq -1$ is a real number. Then,a root of this equation,among the following,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module