NEET Practice Questions (MCQs) with Answers & Solutions

To find the maximum potential difference across the resistor, we use the formula $V_{max} = I_{rms} imes R imes \sqrt{2}$. Given $I_{rms} = 10 ext{ A}$ and $R = 12 \\Omega$, we have $V_{max} = 10 imes 12 imes \sqrt{2} = 169.68 ext{ V}$. Therefore, the maximum potential difference is 169.68 V.

For an AC current, the peak value $I_m$ can be calculated from the RMS value using $I_m = I_{rms} imes ext{√2} = 10 imes ext{√2} = 14.14$ A. The time taken to reach from zero to maximum value for an AC current of frequency 50 Hz is a quarter of the period, i.e., $T/4 = rac{1}{4f} = rac{1}{4 imes 50} = 5 imes 10^{-3}$ seconds.

The power consumed in an AC circuit is given by $P = VI ext{cos}( heta)$ where $θ$ is the phase difference between the voltage and current. Here, $I = I_0 ext{sin}(ωt - rac{π}{2})$ and $E = E_0 ext{sin}(Ωt)$. Since the phase difference is $π/2$, $ ext{cos}(π/2) = 0$. Therefore, the power consumption $P = E_0 I_0 ext{cos}(π/2) = 0$.

To find the RMS value of the given alternating current, we need to use the formula for the RMS value of a function composed of two orthogonal components. Here, the given current is $I = I_1 ext{cos}( ext{wt}) + I_2 ext{sin}( ext{wb})$. The correct formula for the RMS value is $I_{ ext{rms}} = rac{1}{ ext{sqrt 2}} imes ext{sqrt}(I_1^2 + I_2^2)$. Thus, the correct option is $\frac{1}{\sqrt{2}}(I_1^2 + I_2^2)^{\frac{1}{2}}$.

In an AC circuit, power consumption is given by P = V_rms * I_rms * cos(Ï•), where Ï• is the phase difference between the voltage and the current. Here, the current is $I = 5 ext{sin}(100t - \frac{\pi}{2})$ and the voltage is $V = 200 ext{sin}(100t)$. The phase difference Ï• is $\frac{\pi}{2}$. For a phase difference of $\frac{\pi}{2}$, cos(Ï•) = cos($\frac{\pi}{2}$) = 0. Therefore, the power consumption is $0$ watts.

The power dissipated in an AC circuit depends on the phase difference between voltage (V) and current (I). When V and I are in phase, the power is maximized, and when they are out of phase, the power is reduced. This relationship is given by the formula: \( P = VI \cos(\phi) \), where \( \phi \) is the phase angle between V and I.

When the voltage is stepped up n times in the transmission of AC power, the current decreases by n times. Since power loss in transmission lines is proportional to the square of the current (\( P_{loss} = I^2 R \)), the power loss decreases by \( n^2 \) times. Therefore, stepping up the voltage n times decreases the power loss by \( n^2 \) times.

The average value of a sinusoidal voltage over one complete cycle is zero. This is because the positive half-cycle cancels out the negative half-cycle. Mathematically, the integral of a sine function over one complete cycle (from 0 to 2Ï€) is zero.