In an LCR series circuit,R = 18 Omega and th | vedclass

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

Question

(C) The true power $P$ consumed in an $LCR$ series circuit is given by the formula: $P = V_{rms} \cdot I_{rms} \cdot \cos \phi$,where $\cos \phi$ is t

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(C) The true power $P$ consumed in an $LCR$ series circuit is given by the formula: $P = V_{rms} \cdot I_{rms} \cdot \cos \phi$,where $\cos \phi$ is the power factor.We know that the power factor $\cos \phi = \frac{R}{Z}$.Also,the $r.m.s.$ current is given by $I_{rms} = \frac{V_{rms}}{Z}$.Substituting these into the power formula: $P = V_{rms} \cdot \left( \frac{V_{rms}}{Z} \right) \cdot \left( \frac{R}{Z} \right) = \frac{V_{rms}^2 \cdot R}{Z^2}$.Given values: $V_{rms} = 220 \ V$,$R = 18 \ \Omega$,$Z = 33 \ \Omega$.$P = \frac{220^2 \cdot 18}{33^2} = \frac{48400 \cdot 18}{1089}$.$P = \frac{871200}{1089} = 800 \ W$.Therefore,the true power consumed is $800 \ W$.

In an $LCR$ series circuit,$R = 18 \ \Omega$ and the impedance $Z = 33 \ \Omega$. An $r.m.s.$ voltage of $V_{rms} = 220 \ V$ is applied across the circuit. The true power consumed in the $a.c.$ circuit is: (in $W$)

Similar Questions

In the adjoining figure,the impedance of the circuit will be $... \Omega$.

How does the sign of the phase angle $\phi$,by which the supply voltage leads the current in an $LCR$ series circuit,change as the supply frequency is gradually increased from very low to very high values?

What is the impedance of an $L-C-R$ series $AC$ circuit?

For a series $LCR$ circuit,which of the following statements is correct?

In an $LCR$ series circuit, the potential difference across the terminals of the inductor, capacitor and resistor is $60 \text{ V}$, $30 \text{ V}$ and $40 \text{ V}$ respectively. Then the supply voltage will be equal to (in $\text{ V}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module