Difference between AC and DC

Ans. There are circumstances where AC is needed to be converted into DC to fulfil the need for power supply. AC is preferred over DC for the long-distance flow of current. AC is transmitted with the help of a transformer and a further rectifier is used for conversion. Voltage both high and low of AC is converted to a safer magnitude for various uses. This conversion of AC to DC is significant as it prevents the electric current from flowing in reverse.

Ans. The difference between AC and DC is that AC can be readily transformed to high or low voltage levels which is difficult to do with DC. In simpler terms, AC voltage can be changed easily. A higher voltage of AC is more efficient for transmission to great distances, which is not possible with DC. Hence, AC can travel long distances, is less expensive and has lesser power loss during transmission.

Ans. Energy is lost while transmission in transformers due to copper loss as copper wires heat up and that's why thick copper wires with low resistance are preferred. Secondly, eddy current loss which means magnetic flux gives rise to eddy currents in the iron core and causes some energy loss in the form of heat. It can be reduced using a laminated iron core.

Ans. A transformer is an electrical device that converts an alternating current to step up or step down the voltage. They improve the safety and efficiency of power distribution by lowering or raising the voltage whenever needed. It uses the principle of electromagnetic induction as the base of its functioning.

Ans. AC which means alternating current and DC which depicts direct current are usually referred to as voltage. Things that work with the help of a battery or use an AC adapter mainly rely on Direct Current. A few of these appliances are flashlights, electric vehicles, cell phones etc. AC cannot be stored in batteries as they cannot supply power due to changing polarity. DC is used to charge batteries and supply power to many electrical devices. The basic source of direct current is produced by batteries, electrochemical cells, photovoltaic cells etc.

Ans. On decreasing the frequency of the AC source, reactance, xC=1ω/C will increase, which will lead to a decrease in conduction current. In this case
I_D = I_C
Hence, the displacement current will decrease.

Ans. (I) P = V2/R → R = V2/P = 220 X 220/ 100 

= 484 Ω

(ii) I_rms = V_rms/R = 220/484 = 0.45 ampere

Hint: (i) 242Ω
(ii) I_rms = 0.90 atmosphere 

Ans. (i) Increases. X_L = ωL
The number of turns decreases, L decreases, and hence current through the bulb increases. Also voltage across the bulb increases.
(ii) Decreases: Iron rod increases the inductance which increases X_L, hence current through the bulb decreases./Voltage across the bulb decreases.
(iii) Increases. Under this condition (X_C = X_L) the current through the bulb will become maximum.

Ans. (i) On introducing a dielectric slab, brightness will increase due to an increase in capacitance.
(ii) Brightness will decrease, as the resistance (R) is increased, and the potential drop across the bulb will decrease (since both are connected in series).

Ans: Some of the characteristic properties of material suitable for the core of a transformer:

• It should have high permeability
• It should have low hysteresis loss.
• It should have low coercivity/retentivity.
• It should have high resistivity.

Ans. The rectifier converts AC, which periodically reverses direction, into DC, which flows in only one direction.

Ans. The following are the disadvantages of AC

• AC can be more dangerous than DC because the maximum value of DC is √2 times the effective value of e.g. the domestic supply of 220 V has a maximum value given by (√2 x 220)V i.e. 311 V.
• AC cannot be used in electrolysis processes (such as electroplating, electrotyping, and electro-refining, etc.) where only DC is used.
• AC in a wire is not uniformly distributed throughout its cross-section. The AC density is much greater near the surface of the wire than that inside the wire. This concentration of AC near the surface of the wire is called the skin effect. In fact, the inner conducting material of a wire remains unused.

Ans. The electric current that takes a short finite time to grow to maximum value or decay to minimum value in an electric circuit containing an inductor, or a capacitor or both is called Transient current.

Ans. Let an alternating current I = I₀ sin ωt flow through a conductor of resistance R for time dt.

The heat energy produced in the conductor is given by

dH = I²R dt = (I₀ sin ωt )²R dt

⇒ dH = I₀²R sin² ωt dt

Now the heat is produced in the conductor when the current flows for the time period T (i.e. from t = 0 to t = T) is given by

∫ dH = ∫₀^T I₀²R sin²t dt

I ⇒ H = I₀²R∫₀^Tsin²t dt

On solving the above equation, we get

H = \( \frac{I_0^2RT}{2}\) …(i)

Let I_rms be the rms value of AC which flows through the conductor of resistance R for time T.

Therefore, the heat produced in the conductor is given by

H = I_rms ²RT …(ii)

From equation (i) and (ii), we get

I_rms²RT = \( \frac{I_0^2RT}{2}\)

I_rms = \(\frac{I_0}{\sqrt{2}}\) = 0.707 I₀

Ans. Wattless current is that component of current due to which the power consumed in the circuit is zero.

Wattfull current is that component of current due to which power is consumed in the circuit.

Ans. The mean or average value of AC is the value of a steady current that sends the same amount of charge through a circuit in a certain time interval, as is sent by an AC through the same circuit in the half cycle.

The mean value of an AC over half cycle is 63.7% of its peak value.

Ans. The frequency of AC supply in India is 50 Hz, whereas the frequency of DC is 0.

Ans. The equation for the sinusoidal voltage is given as 

V = 200 sin 314t

By comparing the above equation with the standard equation i.e. V = V₀ sin ωt, we get

Peak value of voltage, V₀ = 200 V

1. The relationship between the rms value and peak value of the sinusoidal voltage is given by

V_rms = 0.707 x V₀

On substituting the value of V₀ we get

V_rms = 0.707 x 200 = 141.44 V

2. RMS value of AC, I_rms = V_rms/R

⇒ I_rms = 141.44/10 = 14.144 A

100. The power dissipated as heat is given by

P = V_rmsI_rms

On substituting the values, we get

P = 141.44 x 14.144 = 2000.5 watts