**Solution of GATE Previous Year 2012 Solved paper**

**Q1. A two phase load draws the following phase currents:** \mathbf{I_{1}(t)=I_{m}sin(\omega t -\phi_{1})}, \mathbf{I_{2}(t)=I_{m}cos(\omega t -\phi_{2})}. **These currents are balanced if **\mathbf{\phi_{1}} **is equal to**\\\;\\\;\;\mathbf{(A)\;-\phi_{2}}\;\;\;\;\;\mathbf{(B)\;\phi_{2}\;\;\;\;\;} \mathbf{(C)\;(\frac{\pi}{2}-\phi_{2})\;\;\;\;\;} \mathbf{(D)\;(\frac{\phi}{2} + \phi_{2})}

**Solution:**

\;I_{1}(t)=I_{m}Sin(\omega t - \phi_{1}) = I_{m}cos(90- (\omega t - \phi_{1}))\\\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\because cos(90-\theta)=sin(\theta)\\\;\\\;I_{1}(t)=I_{m}cos(\omega t - \phi_{1}-90)=I_{m}cos(\omega t -(\phi_{1}+90))\\\;\\I_{1}=I_{m}cos(\phi_{1}+90)

I_{2}=I_{m}cos(\omega t - \phi_{2}) = I_{m}cos \phi_{2}\\\;\\Now, in phasor form\\\;\\I_{1}=I_m\angle (90 + \phi_{1}) \\\;\\I_{2}=I_{2}\angle \phi_{2}

Currents will be balanced if \\I_{1} + I_{2} = 0 \\\;\\I_{m}\angle(90+\phi_{1})+I_{m}\angle \phi_{2} = 0

I_{m}cos(90+\phi_{1})+jI_{m}sin(90+\phi_{1}) + I_{m}cos\phi_{2}+jI_{m}sin \phi_{2} = 0\\\;\\I_{m}cos(90+\phi_{1})+I_{m}cos\phi_{2}+j(I_{m}sin(90+\phi_{1})+I_{m}sin\phi_{2}) =0\\\;\\I_{m}(cos(90+\phi_{1})+cos\phi_{2})+jI_{m}(sin(90+\phi_{1})+sin\phi_{2}) =0

cos(90+\phi_{1})+cos\phi_{2} = 0 \\\;\\cos(90+\phi_{1})=-cos\phi_{2}\\\;\\cos(90+\phi_{1})=cos(\pi+\phi_{2}) \\\;\\ 90 + \phi_{1} = \pi + \phi_{2}\\\;\\ \phi_{1}=\frac{\pi}{2}+ \phi_{2}

**Q2.** **The impedance looking into the node 1 and 2 in the given circuit is**\\\;\\\;\;\;\;\;\;\;\;\;

**Solution**: Simplify the diagram as show below\\\;\;\;\;\;

Now Apply voltage test source of 1A between the terminals “1” and “2” as shown below \\\;\;\;\;\;

We will slightly further modify the diagram and name the nodes as shown below.\\\;\;\;\;\;

Now Apply The Kirchhoff’s current law (KCL) at node “a”\\\;\;\;i_b + 99i_b + 1 = i_c\\\;\\\;\;\;100i_b + 1 = i_c\;\;\;\;\;\;\;\;\;\;\;(1)\\\;\\we know \\\;\\\;\;\;i_b = -\frac{V_{test}}{10K}\\\;\\\;\;\;i_c = \frac{V_{test}}{100}

Putting the values of I_{b} and I_{c} into equation (1)\\\;\\\;\;\;-\frac{V_{test}}{100}+1=\frac{V_{test}}{100}\\\;\\\;\;\;1 = \frac{V_{test}}{50}\\\;\\\;\;\;V_{test}= 50

Impedance or Resistance across the node “1” and “2” = \frac{V_{test}}{1}=\frac{50}{1}=50\Omega

Note: solved by using Thevenin theorem

**Q3**: **In the circuit shown below, the current through the inductor is**\\\;\\\;\;\;\;\;\\\;\\\;\;\;\;\;

**Solution:** Let current through 1\Omega is I_{2} as show in figure below\\\;\\\;\;\;\;\;

Applying KVL at node V_{1} \\\;\\1\angle 0 = I_{1}+I_{2}\\\;\\1\angle 0 = \frac{V_{1}-(-1\angle 0)}{j1}+\frac{V_{1}-(-1\angle 0)}{1}

j = V_{1}(1+j) +j + 1\angle 0 \\\;\\ V_{1}= -\frac{1}{1+j}

Now, Find I_{1} \\\;\\ I_{1} = \frac{V_{1}+1\angle 0}{j}\\Putting the Value of V_{1} in above equation \\\;\\I_{1} = \frac{1}{1+j}\\ Thus option “C” is correct.

**Q4: The average power delivered to an impedance (4-j3)\Omega by a current 5cos(100\pi t+100)A is** **\\\;\\ (A) 44.2 W \;\;\; (B) 50W \;\;\; (C) 62.5 W \;\;\; (D) 125W**

**Solution: **Given data \\\;\\Impedance , Z = 4 -j3 \\\;\\Current , I = 5cos(100\pi t+100)A

We Know

Average Power , P=I_{rms}^{2}R = I_{rms}V_{rms}cos\phi \;\;\;\;\;\;\;\;(1)

Impedance, Z = R +jX = 4 – j3 (Where R = Resistance, X = Reactance)\\\;\\Thus R = 4

Current , i = I_{m}cos(\omega t+ \theta) = 5cos(100\pi t + 100)\\\;\\Thus , I_{m} = 5

We know, I_{rms}= \frac{I_{m}}{\sqrt{2}}= \frac{5}{\sqrt{2}}

Putting the Value of I_{rms} and R in equation (1) \\\;\\Average Power, P = (\frac{5}{\sqrt{2}})^2 \times 4 = 50W

**Q5: If V _{A} – V_{B} = 6V, then V_{C} – V_{D} is \\\;\\**

**\\\;\\ (A) -5V \;\;\;\;\;(B) 2V \;\;\;\;\;(C) 3V \;\;\;\;\;(D) 6V**

**Solution: **In the given circuit V_{AB} = V_{A} – V_{B} = 6V\\\;\\ Current in the branch “AB” is I = \frac{V_{A}-V_{B}}{2}= \frac{6}{2}=3A

Consider The open port network as shown below

Current at one terminal is equal to current leaving the other terminal. Thus current in branch “AB” is equal to current in branch “DC”.

Total current in 1\Omega is 5A ( By KCL at node “D”)

Thus V_{C} + 5 – V_{D} = 0 \\\;\\V_{C} – V_{D} = – 5V (By Applying KVL to branch “CD” )

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