Test: AC Applied Across Resistor - NEET MCQ

 Answer :- d

Solution :- as we know that H= (I₀² R)/2 * T/2 -----------------------(1)

If I(rms) be the rms value of ac then

H = I²_(rms)R T/2 ---------------------(2)

From eq (1)&(2)

I²_(rms)R T/2 = (I₀²R)/2 * T/2

I²_(rms) = (I₀²)/2

= I_(rms)= (I₀)/(2)^½

= I_(rms) = 0.707 I₀

Compare the given eqn. with the standard from I=I0sinωt
I₀=8, Irms=I₀/√2=8/√2=5.656A

Time constant is a measure of delay in an electrical circut resulting from either an inductor and resistor or capacitor and resistor. I will discuss rhe most common case which is resistor and capacitor, however the inductor resistor combination behaves in a similar manner. The time constant is equal to the value of the resistance in ohms multiplied by the value of capacitance in Farads. The time constant is measured in seconds . It represents the time for the voltage to decay to 1/2.72.

Answer :- b

Solution :-  f = 75hz

w=2πf

= 2 * π * 75

= 150π

E(max) = (2)^½ E(rms)

E(max) = 1.414 * 200

= 282.8V

E(ins) = E(max)sinwt

E(ins) = 282.8 sin 150πt

We know that,
 ω=(1/2)CV²
After putting the values,
=(1/2)x9.2x22.5x22.5
=2328.75J
Hence option B is the answer.

Alternating current is an electric current which periodically reverses direction, as opposed to direct current which flows only in one direction. And it can be easily represented by the periodic function.
So, I = I_o sin wt or I = l_o cos wt.

peak value E_m=2E₀/π
so 2/π value is 0.637
therefore,
E_m=-0.637 E₀

The only component that dissipates energy in ac circuit is the resistor because  Pure Inductive and pure capacitive circuits have no power loss.

220 volt a.c. means the effective or virtual value of a.c. is 220 volt, i.e., E_v=220
As peak value E0=√2Ev
∴E₀=1.414×220=311 volt
But 220 volt d.c. has the same peak value (i.e., 220 volt only).
Moreover, the shock of a.c. is attractive and that of d.c. is repulsive.
Hence 220 volt a.c. is more dangerous than 220 volt d.c.
 

Power=IV
Where I=V_rms value of current=I_V
And V=V_rms value of voltage=E_V
Therefore, P=E_VI_V