Electric Circuits Test 10

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Electric Circuits Test 10

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Weekly Quiz Competition

Question 1
1 / -0

Given the time domain voltage v = 4 cos ((π/6)t). The expression for instantaneous power across an impedance z = 2 L60^ois:

A
\(8\cos \left( {\frac{\pi }{6}t} \right)\)

B
\(2 + 4\cos \left( {\frac{\pi }{3}t - 60^\circ } \right)\)

C
\(8\cos \left( {\frac{\pi }{3}t - 60^\circ } \right)\)

D
\(2 + 4\cos \left( {\frac{\pi }{6}t - 60^\circ } \right)\)

Solution

\(v = 4\cos \left( {\frac{\pi }{6}t} \right)\)
z = 2 L60^o
Phase current \(I = \frac{V}{Z} = \frac{{4\angle 0^\circ }}{{2\angle 60}} = 2\angle - 60^\circ \)
In time-domain \(I = 2\cos \left( {\frac{\pi }{6}t - 60^\circ } \right)\)
P(t) = V(t)I(t)
\( = \left[ {4\cos \left( {\frac{\pi }{6}t} \right)} \right]\left[ {2\cos \left( {\frac{\pi }{6}t - 60} \right)} \right]\)
\( = 2 + 4\cos \left( {\frac{\pi }{3}t - 60^\circ } \right)\)
Trigonometric multiplication shortcut:
p(t) = v(t) i(t)
p(t) = V_m cos ωt I_m cos (ωt - ϕ)
\(p\left( t \right) = \frac{{{V_m}{I_m}}}{2}\cos \phi + \frac{{{V_m}{I_m}}}{2}\cos \left( {2\omega t - \phi } \right)\)
For this question,
V_m = 4, I_m = 2, ϕ = 60°
\(p\left( t \right) = \frac{{\left( 4 \right)\left( 2 \right)}}{2}\;\left( {\frac{1}{2}} \right) + \frac{{\left( 4 \right)\left( 2 \right)}}{2}\cos \left( {\frac{\pi }{3}t - 60^\circ } \right)\)
\(p\left( t \right) = 2 + 4\cos \left( {\frac{\pi }{3}t - 60^\circ } \right)\)
Question 2
1 / -0

The power consumed by a coil is 300 W when connected to a 20 V dc source and 175 W when connected to a 30 V ac source the reactance of the coil is – (in Ω)

A
2.2-2.3

B
2.24-2.31

C
2.27-2.36

D
2.29-2.32

Solution

P = 300 W, V = 20 DC
\(\frac{{{V^2}}}{R} = 300 \Rightarrow R = \frac{{20 \times 20}}{{300}} = 1.33\;{\rm{\Omega }}\)
P = 175 W, V = 30 AC
⇒ I²R = 175
\(\Rightarrow {I^2} = \frac{{175}}{{1.33}} = 131.25 \Rightarrow I = 11.45\;A\)
V = IZ
⇒ 30 = 11.45 (Z) ⇒ Z = 2.61 Ω
\(\Rightarrow X = \sqrt {{{\left( {2.61} \right)}^2} - {{\left( {1.33} \right)}^2}} = 2.25\;{\rm{\Omega }}\)
Question 3
1 / -0

The voltage across the circuit and the current flows through it, are given by following expressions:
V(t) = 10 – 5 sin (ωt + 60°) V
i(t) = 5 + 10 sin (ωt) A
Which of the following statement is/are true?

A
The real power delivered to the circuit is 12.5 W

B
The reactive power delivered to the circuit is 21.65 W

C
The apparent power of the circuit is 66.14 W

D
The average power delivered to the circuit is 62.5 W

Solution

V(t) = 10 + 5 sin (ωt + 60°) V
i(t) = 5 + 10 sin (ωt) A
Real power (or) average power = V_rms I_rms cos ϕ
= V_rms I_rms cos ϕ
\(= \left( {10} \right)\left( 5 \right) + \left( {\frac{5}{{\sqrt 2 }}} \right)\left( {\frac{{10}}{{\sqrt 2 }}} \right)\cos 60^\circ\)
= 62.5 W
Reactive power = V_rms I_rms sin ϕ
\(= \left( {\frac{5}{{\sqrt 2 }}} \right)\left( {\frac{{10}}{{\sqrt 2 }}} \right)\sin 60^\circ \)
= 21.65 W
Apparent power (s) \(= \left( {10} \right)\left( 5 \right) + \left( {\frac{5}{{\sqrt 2 }}} \right)\left( {\frac{{10}}{{\sqrt 2 }}} \right)\)
= 75 W