1
GATE EE 2012
+1
-0.3
The average power delivered to an impedance $$\left(4-j3\right)$$ Ω by a current $$5\cos\left(100\mathrm{πt}+100\right)\;A$$ is
A
44.2 W
B
50 W
C
62.5 W
D
125 W
2
GATE EE 2012
+1
-0.3
A two phase load draws the following phase currents: $$i_1\left(t\right)=I_m\sin\left(\omega t-\phi_1\right)$$, $$i_2\left(t\right)=I_m\cos\left(\omega t-\phi_2\right)$$. These currents are balanced if $$\phi_1$$ is equal to
A
-$$\phi_2$$
B
$$\phi_2$$
C
$$\left(\mathrm\pi/2\;-\;{\mathrm\phi}_2\right)$$
D
$$\left(\mathrm\pi/2\;+\;{\mathrm\phi}_2\right)$$
3
GATE EE 2011
+1
-0.3
The r.m.s value of the current i(t) in the circuit shown below is
A
$$\frac12A$$
B
$$\frac1{\sqrt2}A$$
C
1 A
D
$$\sqrt2\;A$$
4
GATE EE 2011
+1
-0.3
The voltage applied to a circuit is $$100\sqrt2\cos\left(100\mathrm{πt}\right)$$ volts and the circuit draws a current of $$10\sqrt2\;\sin\left(100\mathrm{πt}+\mathrm\pi/4\right)$$ amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is
A
$$10\sqrt2\;\angle-\mathrm\pi/4$$
B
$$10\;\angle-\mathrm\pi/4$$
C
$$10\;\angle+\mathrm\pi/4$$
D
$$10\sqrt2\;\angle+\mathrm\pi/4$$
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