Difference Between AC and DC: MCQ

Ans. (c) 100/π cycle/sec

Explanation: X_c = \(\frac{1}{\omega C}\)

= \(\frac{1}{2 \pi f C}\)

f = \(\frac{1}{2 \pi X_c C}\)

= \(\frac{1}{2 \pi \ \times \ 1000 \ \times \ 5 \ \times \ 10^{-6}}\)

= \(\frac{100}{\pi}\) Hz

Ans. (b) Admittance

Explanation: When a combination of resistance (inductance or resistance) and capacitance or all three jointly are connected in a circuit, then the opposition to the current in this circuit by this combination is known as the impedance of the circuit. Thus, the reciprocal of impedance is termed as the admittance of the AC circuit.

Ans. (c) 5×10^-3 sec and 14.14 amp

Explanation: As per the given question, the time taken by Alternating current in order to reach maximum from zero is \(\frac{T}{4}\) (T = time period of AC function)

Thus, T =\(\frac{1}{f}\), Frequency (f) = 50 Hz.

Now, the time is taken to reach the maximum = \(\frac{T}{4} = \frac{1}{4f} = \frac{1}{4 \ \times \ 50} = 5 \times 10^{-3} \ sec\)

I_RMS = \(\frac{I_0}{{\sqrt 2}}\), I₀ = Peak value of Current

Thus, the Peak Value of current I₀ = \(\sqrt{2} \ \times \ I_RMS = \ \sqrt{2} \times 10 =14.14 \ A\)

Ans. (c) The average value of a complete cycle is zero

Explanation: The reason why alternating current cannot be measured by the D.C. ammeter is that the A.C. current usually has alternating signs, thus the average value of current for the complete cycle is zero. Herein, the practical frequencies are roughly about 50-60 Hz, meaning that it switches signs over 100 times a second which possibly is not visible on D.C. ammeters.

Ans. (b) 5 V

Explanation: Assume that V₁, V_2,​ and V₃​ are the voltages attached across resistance, inductance, and capacitance.

Hence, the source voltage (V) can be considered as:

V =\(\sqrt{v{^2_1} +(v_3 - v_2)^2} = \sqrt{5^2 + (10 - 10)^2}\)

= 5 V

Ans. (a) Fixed Polarity

Explanation: The direction and magnitude of the current, in a Direct Current (DC), do not change. Simply, both positive and negative terminals of a battery are always positive and negative. Therefore, the current that flows always is in the same direction between both terminals. Examples: Fuel cells, Batteries, and Solar cells.

Ans. (b) 220 V

Explanation: Potential difference can be expressed as the difference between two points in a circuit in the amount of energy that the charge carriers have. The potential difference between a live wire and a neutral wire is 220 V. An electric fuse is usually located in the phase wire’s path before it is attached to the electric meter.

Ans. (a) 0.33 mA

Explanation: The amount of current gain, (α) = 0.967

Emitter current = 10 mA

Now, in order to find the base current, we have to:

The common gain DC current us is given by,

α = 0.967 = I_E/​I_C ​ ​=10/I_C​​

I_C​ = 0.967∗10

I_C = 9.67mA.

The base current of the transistor can be expressed by the following formula,

I_E​ = I_B​ + IC​

10 = I_B​ + 9.67

I_B​ = 0.33mA.

Thus, the value of the base current of the transistor is 0.33mA.

Ans. (d) \(\frac{1}{2}Ip^2R\)

Explanation: Value of r.m.s current, I_max​ = \(\frac{I_p}{\sqrt{2}}\)

So power dissipated is, P = \(I^2_{rms}R = \frac{1}{2}I^2_{p}R\)

Thus, the answer is \(\frac{1}{2}Ip^2R\).

Ans. (b) The electrolyte used in the cell

Explanation: When electric current passes via a cell, the value of resistance that is mainly offered by the electrodes and electrolyte is known as the internal resistance of a cell.

Ans. (a) 150 Hz and 14.14 A

Explanation: Given, I = 20 sin 30πt, comparing it with I = I₀ sinωt, we get

I₀ = 20 and ω = 300π

Now, frequency, f = ω/2π = 300π/2π = 150 Hz

I_rms = I₀/\(\sqrt 2\) = 20/1.414 = 14.14 A

Ans. (c) 424.3 V

Explanation: Given, the peak value of alternating voltage, V₀ = 600 V

We have, rms voltage, V_rms = V₀/\(\sqrt 2\) = 600/1.414 = 424.3 V