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Answer : d
Solution : as we know that H= (I_{0}^{2 }R)/2 * T/2 (1)
If I(rms) be the rms value of ac then
H = I^{2}_{(rms)}R T/2 (2)
From eq (1)&(2)
I^{2}_{(rms)}R T/2 = (I_{0}^{2}R)/2 * T/2
I^{2}_{(rms)} = (I_{0}^{2})/2
= I_{(rms)}= (I_{0})/(2)^{½}
= I_{(rms)} = 0.707 I_{0}
Given the instantaneous value of current from a.c. source is I = 8 sin 623t. Find the r.m.s value of current
Compare the given eqn. with the standard from I=I0sinωt
I_{0}=8, Irms=I_{0}/√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.
Find the instantaneous voltage for an a.c. supply of 200V and 75 hertz
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
If a capacitor of capacitance 9.2F has a voltage of 22.5V across it. Calculate the energy of the capacitor.
We know that,
ω=(1/2)CV^{2}
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.
What is the relationship between E_{m} and E_{0}
peak value E_{m}=2E_{0}/π
so 2/π value is 0.637
therefore,
E_{m}=0.637 E_{0}
The only component that dissipates energy in ac circuit is:
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
∴E_{0}=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.
The average power dissipation in pure resistive circuit is:
Power=IV
Where I=V_{rms} value of current=I_{V}
And V=V_{rms} value of voltage=E_{V}
Therefore, P=E_{V}I_{V}
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