Question 81:
easyPower dissipated in an LCR series circuit connected to an AC source of emf $varepsilon$ is:
(2009)
Average power dissipated is $P = V_{rms} I_{rms} \cos\phi$.
$P = \varepsilon \left(\frac{\varepsilon}{Z}\right) \left(\frac{R}{Z}\right) = \frac{\varepsilon^2 R}{Z^2}$.
In a series LCR circuit, $Z^2 = R^2 + (L\omega - \frac{1}{C\omega})^2$.
Thus, $P = \frac{\varepsilon^2 R}{[ R^2 + (L\omega - \frac{1}{C\omega})^2 ]}$.
Question 82:
easyIn an AC circuit the e.m.f (e) and the current (i) at any instant are given respectively by $e = E_0 sin(\omega t)$, $i = I_0 sin(\omega t – phi)$. The average power in the circuit over one cycle of a.c. is:
(2008)
Average power is given by $P_{avg} = E_{rms} I_{rms} \cos\phi$.
Substituting RMS values: $E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.
$P_{avg} = \left(\frac{E_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos\phi$.
$P_{avg} = \frac{E_0 I_0}{2} \cos\phi$.
Question 83:
easyA coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. It is placed in series with a condenser of $25 \Omega$. The combination is connected to an AC source of $110 V$. The power factor of the circuit is:
(2006)
Net reactance is $X = X_L - X_C = 31 - 25 = 6 \Omega$.
Impedance $Z = \sqrt{R^2 + X^2} = \sqrt{8^2 + 6^2} = 10 \Omega$.
Power factor is $\cos\phi = \frac{R}{Z}$.
$\cos\phi = \frac{8}{10} = 0.80$.
Question 84:
easyFor a series LCR circuit the power loss at resonance is:
(2002)
At resonance in a series LCR circuit, $X_L = X_C$, meaning the circuit is purely resistive.
Total impedance $Z = R$.
Power loss is $P = I_{rms}^2 Z \cos\phi$.
Since $\cos\phi = 1$ and $Z = R$, $P = I^2 R$.
Question 85:
easyIn an a.c. circuit with phase voltage V and current I, the power dissipated is
(1997)
The average power dissipated in an A.C. circuit is $P = V_{rms} I_{rms} \cos\phi$.
Here, $\phi$ is the phase angle between the voltage and the current.
The term $\cos\phi$ is known as the power factor.
Therefore, power dissipated depends on the phase angle between V and I.
Question 86:
easyIn an A.C. circuit, the current flowing is $I = 5 \sin (100t – pi/2)$ ampere and the potential difference is $V = 200 \sin (100t)$ volts. The power consumption is equal to
(1995)
The phase difference between voltage $V = 200 \sin(100t)$ and current $I = 5 \sin(100t - \pi/2)$ is $\phi = \pi/2$.
Power factor of the circuit is $\cos\phi = \cos(\pi/2) = 0$.
Average power consumption is $P = V_{rms} I_{rms} \cos\phi$.
$P = V_{rms} I_{rms} \times 0 = 0 W$.
Question 87:
easyTwo cables of copper are of equal lengths. One of them has a single wire of area of cross-section A, while other has 10 wires of cross-sectional area A/10 each. Give their suitability for transporting A.C. and D.C
(1994)
Alternating current (A.C.) exhibits the skin effect, concentrating on the conductor's surface.
Multiple stranded wires have a larger effective surface area, reducing resistance for A.C.
Direct current (D.C.) flows uniformly across the entire cross-section.
Thus, multiple strands are suitable for A.C., and either can be used for D.C.