Q21GATE 2014NAT

A Y-network has resistances of 10 $\Omega$ each in two of its arms, while the third arm has a resistance of 11 $\Omega$ . In the equivalent $\Delta$ -network, the lowest value (in $\Omega$ ) among the three resistances is _____.

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📖 Explanation

According to the question The equivalent $\triangle$ -network of the above Y-network is

\begin{aligned} \text { Here, } R_{a} &=10+11+\frac{10 \times 11}{10}=32 \Omega \\ R_{b} &=10+11+\frac{10 \times 11}{10}=32 \Omega \\ R_{c} &=10+10+\frac{10 \times 10}{11}=29.09 \Omega \end{aligned}

Hence, the lowest value among the three resistances is $29.09 \Omega$

Q23GATE 2014MCQ

Consider the configuration shown in the figure which is a portion of a larger electrical network For R = 1 $\Omega$ and currents $i_{1} = 2 A$ , $i_{4} =- 1A$ , $i_{5}=- 4 A$ , which one of the following is TRUE ?

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📖 Explanation

Given data:

\begin{array}{l} i_{1}=2 \mathrm{A}, i_{4}=-1 \mathrm{A}, i_{5}=-4 \mathrm{A} \\ R=1 \Omega \end{array}

To calculate: $i_{6}=?$ Using KVL at all the three nodes we get, At node A $i_{5}-i_{3}+i_{2}=0\qquad \ldots(i)$ At node B $i_{4}+i_{1}-i_{2}=0\qquad \ldots(ii)$ At node C $i_{6}+i_{3}-i_{1}=0 \qquad \ldots(iii)$ By putting the value of $i_{3}$ and $i_{2}$ from equation (i) and (ii) in equation (iii) we get,

\begin{aligned} i_{6}+\left(i_{2}+i_{5}\right)-i_{1}&=0 \\ i_{6}+\left(i_{1}+i_{4}+i_{5}\right)-i_{1}&=0 \\ \therefore \quad i_{6}+(2-1-4)-2&=0 \\ i_{6}&=5 \mathrm{A} \end{aligned}

Q24GATE 2014NAT

For the Y network shown in the figure, the value of $R_{1}$ (in $\Omega$ ) in the equivalent $\Delta$ -network is____.

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📖 Explanation

Using star-delta conversions, The value of $R_{1}$ is given by $=5+3+\frac{5 \times 3}{7.5}=10$

Q25GATE 2014MCQ

The circuit shown in the figure represents a

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Q26GATE 2013MCQ

Consider the following figure The current in the $1\Omega$ resistor in Amps is

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📖 Explanation

The current in the $1 \Omega$ resistor $I_{1}=\frac{10}{1}=10 \mathrm{Amp}$

Q27GATE 2013MCQ

Three capacitors $C_{1}$ , $C_{2}$ and $C_{3}$ whose values are 10 $\mu$ F, 5 $\mu$ F, and 2 $\mu$ F respectively, have breakdown voltages of 10V, 5V, and 2V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in $\mu$ C stored in the effective capacitance across the terminals are respectively,

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📖 Explanation

\begin{aligned} Q &=C V \\ Q_{1} &=C_{1} V_{1}=10 \times 10^{-6} \times 10=100 \mu \mathrm{C} \\ Q_{2} &=C_{2} V_{2}=5 \times 10^{-6} \times 5=25 \mu \mathrm{C} \\ Q_{3} &=C_{3} V_{3}=2 \times 10^{-6} \times 2=4 \mu \mathrm{C} \end{aligned}

Capacitors $C_{2}$ and $C_{3}$ are in series. In series charge is same. So, the maximum charge on $C_{2}$ and $C_{3}$ will be minimum of $\left(Q_{2}, Q_{3}\right)=\min (25 \mu C, 4 \mu C)=4 \mu C=Q_{23}$ In series the equivalent capacitance of $C_{2}$ and $C_{3}$ is $C_{23}=\frac{C_{2} C_{3}}{C_{2}+C_{3}}=\frac{5 \times 2}{5+2}=\frac{10}{7} \mu \mathrm{F}$ So, the equivalent voltage $V_{23}=\frac{Q_{23}}{C_{23}}=\frac{4 \times 10^{-6}}{\frac{10}{7} \times 10^{-6}}=\frac{28}{10}=2.8 \mathrm{V}$ In parallel, the voltage is same. $V_{1}=V_{23}=2.8 \mathrm{V}$ Charge in capacitor $C_{1}$ $Q_{1}=C_{1} V_{1}=10 \times 10^{-6} \times 2.8=28 \mu \mathrm{C}$ In parallel, the total charge

\begin{array}{l} Q=Q_{1}+Q_{23} \\ Q=4+28 \\ Q=32 \mu \mathrm{C} \end{array}

Q28GATE 2013MCQ

The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage $V_{WX1}= 100V$ is applied across WX to get an open circuit voltage $V_{YZ1}$ across YZ. Next, an ac voltage $V_{YZ2}= 100V$ is applied across YZ to get an open circuit voltage $V_{WX2}$ across WX. Then $\frac{V_{YZ1}}{V_{WX1}}$ , $\frac{V_{WX2}}{V_{YZ2}}$ are respectively,

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📖 Explanation

\begin{aligned} V_{Y Z_{1}} &=100 \times 1.25 \times 0.8 \\ &=100 \mathrm{V} \end{aligned}

In second case, when 100 V is applied at YZ terminals, this whole 100 V will appear across the secondary winding. Hence,

\begin{aligned} V_{\mathrm{w} x_{2}} &=\frac{100}{1.25}=80 \mathrm{V} \\ \Rightarrow \quad \frac{Y_{\mathrm{YZ}_{1}}}{Y_{\mathrm{wx}_{1}}} &=\frac{100}{100}, \frac{V_{\mathrm{w} \times 2}}{V_{\mathrm{YZ}_{2}}}=\frac{80}{100} \end{aligned}

Q29GATE 2013MCQ

Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k>0, the elements of the corresponding star equivalent will be scaled by a factor of

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📖 Explanation

\begin{aligned} R_{A} &=\frac{R_{b} R_{c}}{R_{a}+R_{b}+R_{c}} \\ R_{a}^{\prime} &=k R_{a} \\ R_{b}^{\prime} &=k R_{b} \\ R_{c}^{\prime} &=k R_{c} \\ R_{A}^{\prime} &=\frac{k R_{b} \cdot k R_{c}}{k R_{a}+k R_{b}+k R_{c}}=\frac{k^{2} R_{b} R_{c}}{k\left(R_{A}+R_{b}+R_{c}\right)} \\ &=k \times \frac{R_{b} R_{c}}{R_{a}+R_{b}+R_{c}} \\ R_{A}^{\prime} &=k R_{A} \end{aligned}

Q30GATE 2013MCQ

Consider the following figure The current $I_{s}$ in Amps in the voltage source, and voltage $V_{s}$ in Volts across the current source respectively, are

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📖 Explanation

Voltage across $1 \Omega$ resistance, $V_{1}=10 \mathrm{V}$ Current through $1 \Omega$ resistance, $I_{1}=\frac{10}{1}=10 \mathrm{Amp}$ Voltage across $2 \Omega$ resistance, $V_{2}=10 \mathrm{V}$ Current through $2 \Omega$ resistance, $I_{2}=\frac{10}{2}=5 \mathrm{Amp}$ Apply KCL at node A

\begin{array}{c} -2+I_{s}+I_{2}+I_{1}=0 \\ I_{s}=2-I_{1}-I_{2}=2-10-5 \\ I_{s}=-13 \mathrm{Amp} \end{array}

Voltage at node A,

\begin{aligned} V_{A} &=10 \mathrm{V} \\ V_{s}-10 &=10 \mathrm{V} \\ V_{s} &=10+10=20 \mathrm{V} \\ V_{s} &=20 \mathrm{V} \end{aligned}

Q31GATE 2012MCQ

The average power delivered to an impedance (4 - j3) $\Omega$ by a current 5cos(100 $\pi$ t +100) A is

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📖 Explanation

Average power is same as RMS power.

\begin{array}{l} P=I_{\mathrm{rms}}^{2} R=\left(\frac{5}{\sqrt{2}}\right)^{2} \times 4 \\ =\frac{25}{2} \times 4=50 \mathrm{W} \end{array}

Note: Power is consumed only by resistance i.e. by real part of impedance.

Q32GATE 2012MCQ

In the circuit shown below, the current through the inductor is

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📖 Explanation

According to KCL at node D there will be no current in voltage sources. According to KCL at node A current through inductor will be $i_{1}=i+1\qquad \ldots(1)$ Applying KVL in loop ACDBA we have

\begin{aligned} 1 \times i+(i+1) j 1+1 \angle 0-1 \angle 0 &=0 \\ i+(i+1) j &=0 \\ (1+j) i &=-j \\ i &=\frac{-j}{1+j} &\ldots(2) \end{aligned}

Therefore from (1) and (2) we have

\begin{aligned} i_{1}&=i+1=\frac{-j}{j+1}+1\\ i_{1}&=\frac{1}{1+j} \end{aligned}

Q33GATE 2012MCQ

If $V_{A}-V_{B}$ =6V, then $V_{C}-V_{D}$ is

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📖 Explanation

\begin{aligned} i &=\frac{V_{A}-V_{B}}{2}=\frac{6}{2} \\ &=3 \mathrm{A}\quad \ldots(i) \end{aligned}

KCL at node B, we have

\begin{aligned} i &=i_{2}+i_{1} \\ i_{2}+i_{1} &=3 \mathrm{A} \quad \ldots(ii) \end{aligned}

KCL at node E , we have $i_{1}=i_{3}+i_{4}\quad \ldots(iii)$ KCL at node D we have

\begin{array}{l} i_{5}=i_{2}+i_{3}+i_{4} \\ i_{5}=i_{2}+i_{1} \\ i_{5}=3 \mathrm{A} \end{array}

KCL at node F , we have

\begin{aligned} i_{6}+2+i_{5}&=0\\ i_{6}&=-2-i_{5} \\ i_{6}&=-5 \mathrm{A}\\ \text{SO} \quad V_{C}-V_{D}&=1 \times i_{6}=-5 \mathrm{V} \end{aligned}

Q34GATE 2011MCQ

In the circuit shown below, the current I is equal to

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📖 Explanation

Converting delta into star, the circuit can be redrawn as below: Equivalent impedance of the circuit,

\begin{aligned} z&=(2+j 4) \|(2-j 4)+2 \\ z&=\frac{(2+j 4)(2-j 4)}{2+j 4+2-j 4}+2 \\ \Rightarrow \quad z&=\frac{4+16}{4}+2=7 \Omega \end{aligned}

Therefore, Eurrent, $I=\frac{14 \angle 0^{\circ}}{7}=2 \angle 0^{\circ} \mathrm{A}$

Q35GATE 2010MCQ

In the circuit shown, the power supplied by the voltage source is

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📖 Explanation

Applying KVL in outer loop,

\begin{aligned} (3+i) 2+(2+i) 2&=10 \\ \Rightarrow \qquad 6+2 i+4+2 i&=10 \\ \Rightarrow \qquad 4 i&=0 \\ \Rightarrow \qquad i&=0 \end{aligned}

Power supplied by the voltage source,

\begin{aligned} P &=V_{i} \\ &=10 \times 0=0 \mathrm{W} \end{aligned}

Q36GATE 2009MCQ

A fully charged mobile phone with a 12 V battery is good for a 10 minute talktime. Assume that, during the talk-time the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?

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📖 Explanation

\begin{aligned} P &=V I \\ \text { Energy } &=P \cdot t=V \cdot|t=(V . t)| \\ I &=2 \mathrm{A} \quad(\text { given }) \\ V \cdot t &=\text { Area under } V-t \text { curve } \\ V \cdot t &=\left(\frac{1}{2} \times 2 \times 600\right)+(10 \times 600) \\ &=600+6000 \\ V \cdot t &=6600 \\ E &=(6600) \times 2=13200=13.2 \mathrm{kJ} \end{aligned}

Q37GATE 2009MCQ

In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power. Which of the following can be the value of the current source I ?