Alternating Current MCQs for NEET — Physics Questions with Answers

fact

The transformer works on the principle of mutual induction and the voltage transformation ratio is given by the formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Here, \(V_p\) is the primary voltage, \(V_s\) is the secondary voltage, \(N_p\) is the number of primary turns, and \(N_s\) is the number of secondary turns. Given \(V_p = 20V\), \(N_p = 500\), and \(N_s = 5000\), the secondary voltage \(V_s\) can be calculated as: \[ V_s = V_p \times \frac{N_s}{N_p} = 20V \times \frac{5000}{500} = 200V \] The frequency remains the same at 50 Hz. Hence, the correct answer is

200V, 50Hz

To find the current in the primary coil, we first need to determine the current in the secondary coil. The power consumed by the bulb is 30W and the voltage across it is 6V. Using the formula \( P = V \times I \), we get: \[ I_s = \frac{P}{V} = \frac{30W}{6V} = 5A \] The transformer is step-down, so the primary voltage is higher than the secondary. The current transformation ratio is given by \( \frac{I_s}{I_p} = \frac{V_p}{V_s} \). Given \(V_p = 200V\) and \(V_s = 6V\), we can calculate the primary current \(I_p\) as: \[ I_p = I_s \times \frac{V_s}{V_p} = 5A \times \frac{6V}{200V} = 0.15A \] Hence, the current in the primary is 0.15 A.

A DC ammeter measures the average value of current. For an alternating current (AC), the average value over a complete cycle is zero because AC alternates in direction and spends equal time in positive and negative half-cycles. Hence, a DC ammeter cannot measure AC.

In AC, the resistance of a coil is influenced by both its inherent resistance and reactance (inductive or capacitive). The reactance adds to the resistance, resulting in a higher effective resistance when compared to DC. This is why the resistance will increase in AC.

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.