Alternating Current Test - 16

As given that,v = 50 Hz, \(I_{rms}\) = 5A

t = \(1\over300\)s

As we know that \(I_{rms}\) = \(I_0\over \sqrt2\)

\(I_0\) = Peak value = \(\sqrt2.I_{rms}\) = \(\sqrt2\) x 5

\(I_0\) = \(5\sqrt2A\)

at, t = \(1\over300\)sec, I = \(I_0\) sin\(\omega\)t = 5\(\sqrt2\) sin 2\(\pi\)vt

= 5\(\sqrt2\) sin 2\(\pi\) x 50 x \(1\over300\)

I = 5\(\sqrt2\) sin \(\pi\over 3\) = 5\(\sqrt2\) x \(\sqrt3\over2\) = 5\(\sqrt{3\over2}\) Amp  \(\Big(\therefore\,\, sin{\pi\over3}={\sqrt3\over2}\Big)\)

I = \({5\sqrt3\over\sqrt2}A\)

To deliver maximum power from the generator to the load, total internal reactance must be equal to conjugate of total external reactance.

So, \(X_{int}=X_{ext}\)

\(X_g=(X_L)=-X_L\)

Hence, \(X_L=-X_g\) (Reactance in external circuit).

As we know that,

The voltmeter in AC reads rms values of voltage

\(I_{rms}\) = \(\sqrt2I_0\) and \(V_{rms}=\sqrt2v_0\)

The voltmeter in AC circuit connected to AC mains reads mean value (<\(v^2\)>) and is calibrated in such a way that it gives rms value of (<\(v^2\)>), which is multiplied by form factor \(\sqrt2\) to give rms value \(V_{rms}\).

As we know that,

The resonant frequency in an L-C-R series circuit is

\(v_0={1\over2\pi\sqrt{LC}}\)

So, to reduce \(v_0\) either increase L or increase C. To increase capacitance, another capacitor must be connect in parallel with the first capacitor.

As we know that, Quality factor (Q) of an L-C-R circuit must be higher so Q is

\(Q={1\over R}\sqrt{L\over C}\)

where R is resistance, L is inductance and C is capacitance of the circuit.

So, for higher Q, L must be large, and C and R should be low.

Hence, option (C) verifies it.

As given that,

\(X_L\) = 1\(\Omega\), R = 2\(\Omega\), \(E_{rms}\) = 6V, \(P_{av}\) = ?

The average power dissipated in the L, R, series circuit with AC source

Then \(P_{av}\) = \(E_{rms}\)\(I_{rms}\) cos \(\phi\)       ...(i)

\(I_{rms}\) = \(I_o\over \sqrt2\) = \(E_{rms}\over Z\)

Z = \(\sqrt{R^2+X_L^2}\) = \(\sqrt{4+1}\) = \(\sqrt5\)

\(I_{rms}\) = \(6\over \sqrt5\)A

cos \(\phi\) = \(R\over Z\) = \(2\over \sqrt5\)

By putting the value of \(I_{rms}\), \(E_{rms}\), cos \(\phi\) in equation (i), then,

\(P_{av}\) = 6 x \(6\over \sqrt5\) x \(2\over \sqrt5\) = \(72\over \sqrt5 \sqrt5\)

= \(72\over5\) = 14.4 watt

As given that,

Secondary voltage \((V_S)\) is:

\(V_S\) = 24 Volt

Power associated with secondary is:

\(P_S\) = 12 Watt

As we know that \(P_S\) = \(V_S\)\(I_S\)

\(I_S\) = \(P_S\over V_S\) = \(12\over 24\) = \(1\over2\)A = 0.5 Amp

Peak value of the current in the secondary

\(I_0=I_S\sqrt2=0.5\sqrt2\)

= \({5\over10}\sqrt2\Big[I_0={1\over \sqrt2}Amp\Big]\)

As per question on increasing frequency, the current increases.

So, the reactance of the circuit must be decreases as increasing frequency.

For a capacitive circuit, when f is frequency of the AC circuits.

\(X_C={1\over \omega C}={1\over 2\pi fC}\)

Clearly when frequency increases, \(X_C\) decreases, to increase current capacitors

For R-C circuit, X = \(\sqrt{R^2+({1\over \omega C})^2}\)

So, option (c) and (d) are verify.

Power loss due to transmission lines having resistance (R) and rms current \(I_{rms}\) is

Power loss = \(I^2_{rms}\)R

So to decrease power loss \(I_{rms}\) and R must be lowers for a constant power supply and energy transmit over large distances at high alternating voltages, so current flowing through the wires will be low because for a given power (P).

So, P = \(E_{rms}\)\(I_{rms}\), \(I_{rms}\) is low when \(E_{rms}\) is high.

Power loss = \(I^2_{rms}\)R = low   (\(\because\) \(I_{rms}\) is low)

Now, at the receiving end high voltage is reduced by using step-down transformers.

For step-up transformer,

Output power = Input power

\(V_SI_S\) = \(V_PI_P\)   (\(\because\) \(V_S>V_P\) so \(I_P\geq I_S\))

So, loss of power during transmission become lower.

As given that:

\(P=I^2Zcos\phi\)      ...(i)

where I is the current, Z = Impedance, cos\(\phi\) power factor

As we know that,

\(P=I^2R\)        ...(ii)

Compare (i) & (ii) equation,

So, (power factor)

(cos\(\phi\) = \(R\over Z\))

where R > 0 and Z > 0

So, cos\(\phi\) = \(R\over Z\) is +ve,

cos\(\phi\) > 0

P > 0.

In house, we are using AC supply, so AC currents are used which are having zero average value over a cycle. In houshald circuit L and C are connected, so R, Z cannot be equal.

So, the line is having some resistance so power factor

\(cos\phi={R\over Z}\neq 0\)

So, \(\phi\neq {\pi\over2}\) ⇒ \(\phi < {\pi\over2}\)

Hence, phase angle of voltage & current lies between 0 and \(\pi\over2\).