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Test: AC Power Analysis- 2 - Question 1

A non-alternating periodic waveform is shown below, It’s from factor is

Detailed Solution for Test: AC Power Analysis- 2 - Question 1

Here, time period is, T = 20 ms; between 10 ms < t < 20 ms, i(t) = 0 A.

Also,

= 50 x 0.01 =0.5 A

Test: AC Power Analysis- 2 - Question 2

The voltage and current in an a.c, circuit are given by

v = 100 sin 314t; i = 10 sin 314t

The average power in the circuit is

Detailed Solution for Test: AC Power Analysis- 2 - Question 2

∴ P_{avg }= I^{2 }R

Now,

(since v and i in phase)

Test: AC Power Analysis- 2 - Question 3

The expression for instantaneous power dissipated in a resistor R connected across a sinusoidal alternating voltage source represented as v = V_{m} sin ωt is given by

Detailed Solution for Test: AC Power Analysis- 2 - Question 3

∴ instantaneous power,

Test: AC Power Analysis- 2 - Question 4

The power delivered by the two voltage sources of 6 V and 24 V as shown below will be

Detailed Solution for Test: AC Power Analysis- 2 - Question 4

Since V_{1} < V_{2}, therefore the current wil! flow from V_{2} to V_{1} as shown below.

Hence, the whole power in the circuit will be supplied by V_{2} only. Total power delivered by the voltage source V_{2} is

P_{T} = I^{2} x (2 + 4) + 6I

= 3^{2 }x 6 + 6 x 3 = 54 + 18

= 72 watts

Test: AC Power Analysis- 2 - Question 5

The power loss in watts in the resistor R shown below is

Detailed Solution for Test: AC Power Analysis- 2 - Question 5

V_{R} = (10 + 5 V_{R}) x R

Given, V_{R }= 5 V

So, 5 = (10 + 5 x 5) R

or,

∴ Power loss in R is

Test: AC Power Analysis- 2 - Question 6

The root mean square value of the current wave given by i = (50 + 30 sin t) A will be

Detailed Solution for Test: AC Power Analysis- 2 - Question 6

= 54.30 A

Test: AC Power Analysis- 2 - Question 7

Assertion (A): In an alternating circuit, if the current is (4 - j3) amperes and the impressed voltage is (100 - j50) volts, the reactive power in the circuit is 100 VA.

Reason (R): True power is the real part of (VI).

Detailed Solution for Test: AC Power Analysis- 2 - Question 7

V = (100 - j50) volt,

I = (4 - j3) A

∴ S = VI* = (100 - j50)(4 - j3)*

= (100 - j50)(4 + j3)

= 550 + j100

∴ P = 100 W and Q = 100 VAR

Q is in VAR not in VA. Hence, assertion is not correct.

Also P is the real part of apparent power (S = VI).

Test: AC Power Analysis- 2 - Question 8

In a purely resistive circuit, the average power

P_{av} is______the peak power P_{max}.

Detailed Solution for Test: AC Power Analysis- 2 - Question 8

Instantaneous power in a purely resistive circuit is

Detailed Solution for Test: AC Power Analysis- 2 - Question 9

In a purely inductive circuit, power consumed = Q = VI sin φ. In inductive circuit (having both R and L) power is consumed by resistor R which is P = VI cos φ.

Test: AC Power Analysis- 2 - Question 10

A series circuit containing passive elements has the following current and applied voltage:

The circuit elements

Detailed Solution for Test: AC Power Analysis- 2 - Question 10

Phase difference between v and i is

∴ power factor = cos φ = cos 75° lag

Since, i lags v therefore, the circuit elements may be resistance and inductance (Z = R + jX_{L}) or it may be resistance, inductance and capacitance Z = R + j(X_{L} - X_{C}), so that p.f. becomes lagging.

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