Attempt the ISRO Electronics & Communication (ECE) 2007 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electronics & Communication (ECE) papers
80 questions.
With fixed value capacitor C and variable voltage V across it, the energy stored in the capacitor is
Answer: B. $0.5 CV²$
Explanation: Energy stored in capacitor is given as: Energy = $\frac{1}{2}CV²$
A dc voltage V is applied to a series RL circuit. The steady state current is
Answer: A. V/R
Explanation: The transient current is given by $i(t) = \frac{V}{R}(1 - e^{-tR/L})$. At steady state ($t \rightarrow \infty$), $i(\infty) = \frac{V}{R}(1 - e^{-\infty}) = \frac{V}{R}$
The time-constant of the network shown in the figure is \n\n[Figure needs to be uploaded later]
Answer: A. CR
Explanation: [Figure needs to be uploaded later]\n$R_{eq} = \frac{R}{2}$, $C_{eq} = 2C$. Time-constant $t = R_{eq}C_{eq} = \frac{R}{2} \cdot 2C = RC$.
In the ideal Op-amp circuit shown, $V_0$ is \n\n[Figure needs to be uploaded later]
Answer: C. $3 V_s$
Explanation: [Figure needs to be uploaded later]\n$V_0 = (1 + \frac{R_f}{R_1})V_s = (1 + \frac{2R}{R})V_s = 3V_s$
If the unit step response of a system is a unit impulse function, then the transfer function of such a system will be
Answer: C. s
Three resistors of 6 ohms each are connected as shown in the following figure. The equivalent resistance between $X_1$ and $X_2$ is \n\n[Figure needs to be uploaded later]
Answer: A. 2 ohms
Explanation: [Figure needs to be uploaded later]\nThese three resistances are in parallel, so $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6}$. $R_{eq} = \frac{6}{3} = 2 \Omega$.
The source in the circuit shown is a sinusoidal source. The supply voltages across various elements are marked in the figure. The input voltage is \n\n[Figure needs to be uploaded later]
Answer: B. 5 V
Laplace transform of $e^{-at} f(t)$ is
Answer: C. $F(s+a)$
$\cos \theta$ can be represented by
Answer: C. $\frac{e^{j\theta}+e^{-j\theta}}{2}$
Of the following transfer functions of second order linear time-invariant systems, the underdamped system is represented by
Answer: D. $H(S)=\frac{1}{S²+3S+4}$
Explanation: Comparing $H(s)=\frac{1}{s²+3s+4}$ with standard form $H(s)=\frac{\omega_n^2}{s²+2\xi\omega_ns+\omega_n^2}$, we get $2\xi\omega_n=3$ and $\omega_n=2$. Thus, $\xi=\frac{3}{4}=0.75$. Since $\xi<1$, the system is underdamped.
A differential amplifier has a differential gain of 20,000. CMRR = 80 dB. The common mode gain is given by
Answer: A. 2
Explanation: $CMRR_{dB} = 20 \log(\frac{A_d}{A_{cm}})$. $80 = 20 \log(\frac{A_d}{A_{cm}})$. $\frac{A_d}{A_{cm}} = 10^4 = 10000$. $A_{cm} = \frac{A_d}{10000} = \frac{20000}{10000} = 2$.
Two bulbs marked 200 watt - 250 volts and 100 watt - 250 volts are joined in series to 250 volt supply. Power consumed in circuits is
Answer: B. 67 watt
Explanation: [Figure needs to be uploaded later]\n$P_1 = \frac{V_1^2}{R_1} \Rightarrow 200 = \frac{250^2}{R_1} \Rightarrow R_1 = 312.5 \Omega$. $P_2 = \frac{V_2^2}{R_2} \Rightarrow 100 = \frac{250^2}{R_2} \Rightarrow R_2 = 625 \Omega$. $R_{eq} = R_1 + R_2 = 937.5 \Omega$. $I = \frac{V}{R_{eq}} = \frac{250}{937.5} = 0.266 A$. Power consumed $= V \cdot I = 250 \times 0.266 = 66.67 \approx 67$ Watt.
The 6 V zener diode shown in the figure, has zero zener resistance and a knee current of 5 mA. The minimum value of R so that the voltage across it does not fall below 6 V is \n\n[Figure needs to be uploaded later]
Answer: B. 80 ohms
Explanation: [Figure needs to be uploaded later]\n$I = \frac{10 - 6}{50} = 80$ mA. $I_{zk} = 5$ mA. $I_{Lmax} = I - I_{zk} = (80 - 5) = 75$ mA. $R_{Lmin} = \frac{V_L}{I_{Lmax}} = \frac{6 V}{75 mA} = 80 \Omega$.
A zener diode in the circuit shown in the figure below, has a knee current of 5 mA, and a maximum allowed power dissipation of 300 mW. What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage $V_0$ at 6 V? \n\n[Figure needs to be uploaded later]
Answer: C. 10 mA, 55 mA
Explanation: [Figure needs to be uploaded later]\n$P_{zmax} = 300$ mW. $I_{zmax} = \frac{300 mW}{6 V} = 50$ mA. $I = \frac{9 - 6}{50} = 60$ mA. $I_{Lmin} = I - I_{zmax} = 60 - 50 = 10$ mA. $I_{zmin} = 5$ mA. $I_{Lmax} = I - I_{zmin} = 60 - 5 = 55$ mA.
The voltages at $V_1$ and $V_2$ of the arrangement shown in figure will be respectively \n\n[Figure needs to be uploaded later]
Answer: D. 6 V and 6 V
Explanation: [Figure needs to be uploaded later]\nIn the first case $D_1$ diode will be forward biased so $V_2 = 6V$. In the second case due to 10 k$\Omega$ resistor there will be some voltage drop across this resistor, so diode $D_2$ is also forward biased and voltage $V_1 = 6V$.
A 24 V, 600 mW, Zener diode is to be used for providing a 24 V stabilized supply to a variable load. Assume that for proper Zener action, a minimum of 10 mA must flow through the Zener. If the input voltage is 32 V, what would be the value of R and the maximum load current? \n\n[Figure needs to be uploaded later]
Answer: D. 320 $\Omega$, 15 mA
Explanation: [Figure needs to be uploaded later]\n$V_z = 24~V, P_z = 600~mW, I_{zk} = 10~mA$. \n$I_{zmax} = \frac{P_{zmax}}{V_z} = \frac{600~mW}{24~V} = 25~mA$. \nWhen $I_{zmax}$ occur, no current flow through $I_L$ i.e. $R_L$ is open circuited. \n$25~mA = \frac{32 - 24}{R} \Rightarrow R = \frac{8~V}{25~mA} = 320~\Omega$. \n$I = \frac{32 - 24}{320} = 25~mA$. \nMaximum load current $I_{Lmax} = I - I_{zk} = 25~mA - 10~mA = 15~mA$.
A half-adder can be constructed using two 2-input logic gates. One of them is an AND-gate, the other is
Answer: D. EX-OR
For one of the following conditions, clocked J-K flip-flop can be used as DIVIDE BY 2 circuit where the pulse train to be divided is applied at clock input
Answer: D. J=1, K=1 and the flip-flop should be a negative edge triggered one
Number of comparators needed to build a 6-bit simultaneous A/D converter is
Answer: A. 63
Explanation: Number of comparators used in flash type (simultaneous) A/D converter is $= 2^n - 1 = 2^6 - 1 = 63$. Where, n = number of bits of A/D converter.
The A/D converter used in a digital voltmeter could be (1) successive approximation type (2) Flash converter type (3) Dual slope converter type. The correct sequence in the increasing order of their conversion time taken is
Answer: B. 2, 1, 3
The circuit is a \n\n[Figure needs to be uploaded later]
Answer: D. SR FF
Explanation: [Figure needs to be uploaded later]\nCircuit is SR Latch.
Which of the following binary number is equal to octal number 66.3
Answer: D. 110110.011
Explanation: $(66.3)_8 = (110110.011)_2$
The Boolean expression for the output of the logic circuit shown in the figure is \n\n[Figure needs to be uploaded later]
Answer: B. $Y=\overline{A}\overline{B}+AB+\overline{C}$
Explanation: [Figure needs to be uploaded later]\n$Y = \overline{(A \oplus B)C} = \overline{A \oplus B} + \overline{C} = A \odot B + \overline{C} = \overline{A}\overline{B} + AB + \overline{C}$
For the identity $AB+\overline{A}C+BC=AB+\overline{A}C$, the dual form is
Answer: B. $(\overline{A}+\overline{B})(A+\overline{C})(\overline{B}+\overline{C})=(\overline{A}+\overline{B})(A+\overline{C})$
Explanation: For duality: Step: change AND to OR and vice-versa. Step: complemented to uncomplent and vice-versa. $(\overline{A}+\overline{B})(A+\overline{C})(\overline{B}+\overline{C})=(\overline{A}+\overline{B})(A+\overline{C})$.
The logic circuit given below converts a binary code $Y_1Y_2Y_3$ into \n\n[Figure needs to be uploaded later]
Answer: B. Gray code
A 4-bit presetable UP counter has preset input 0101. The preset operation takes place as soon as the counter reaches 1111. The modulus of the counter is
Answer: B. 10
A 4-bit synchronous counter uses flip-flops with propagation delay time of 25 ns each. The maximum possible time required for change of state will be
Answer: A. 25 ns
If a counter having 10 FFs is initially at 0, what count will it hold after 2060 Pulses
Answer: A. 000 000 1100
The shift register shown in the given figure is initially loaded with the bit pattern 1010. Subsequently the shift register is clocked, and with each clock pulse the pattern gets shifted by one bit position to the right. With each shift, the bit at the serial input is pushed to the left most position (MSB). After how many clock pulses will the content of the shift register become 1010 again? \n\n[Figure needs to be uploaded later]
Answer: B. 7
The combinational logic circuit shown in the given figure has an output Q which is \n\n[Figure needs to be uploaded later]
Answer: B. A+B+C
Explanation: [Figure needs to be uploaded later]\n$Q = \overline{A} \cdot \overline{B} \cdot C + \overline{A}B \cdot 1 + A\overline{B} \cdot 1 + AB \cdot 1$ \n$= \overline{A} \cdot \overline{B}C + \overline{A}B + A(\overline{B} + B)$ \n$= \overline{A} \cdot \overline{B}C + \overline{A}B + A \cdot 1$ \n$= \overline{A} \cdot \overline{B}C + (A + B)$ \n$= \overline{A} \cdot \overline{B}C + A + B$ \n$= (\overline{A} + A)(A + \overline{B}C) + B = 1 \cdot (A + \overline{B}C) + B$ \n$= A + (B + \overline{B})(B + C) = A + 1(B + C)$ \n$= A + B + C$
A sample-and-hold (S/H) circuit, having a holding capacitor of 0.1 nF, is used at the input of an ADC (analog-to-digital converter). The conversion time of the ADC is 1 $\mu$sec and during this time, the capacitor should not lose more than 0.5% of the charge put across it during the sampling time. The maximum value of the input signal to the S/H circuit is 5 V. The leakage current of the circuit should be less than \n\n[Figure needs to be uploaded later]
Answer: D. 2.5 $\mu$A
Explanation: [Figure needs to be uploaded later]\n$Q=CV=0.1\times 10^{-9}\times 5=5\times 10^{-10}$ C.\n0.5% of $Q = 5\times 10^{-10}\times \frac{0.5}{100} = 2.5\times 10^{-12}$ C.\n$i = \frac{\Delta Q}{\Delta t} = \frac{2.5\times 10^{-12}}{1\times 10^{-6}} = 2.5~\mu A$.
Determine the output voltage of a network shown in figure if the digital input is 1011 \n\n[Figure needs to be uploaded later]
Answer: D. -6.875 V
Explanation: [Figure needs to be uploaded later]\n$V_0 = -R_f(\frac{V_A}{800k} + \frac{V_B}{400k} + \frac{V_C}{200k} + \frac{V_D}{100k})$\n$= -(\frac{5}{8} + \frac{5}{4} + \frac{0}{2} + \frac{5}{1})$\n$= -(0.625 + 1.25 + 0 + 5) = -6.875 V$.
A memory system of size 16 K bytes is required to be designed using memory chips, which have 12 address lines and 4 data lines each. The number of such chips required to design the memory system is
Answer: C. 8
Explanation: Number of chips $= \frac{16\times 1024\times 8}{2^{12}\times 4} = 8$.
In time division multiplexing
Answer: C. Total time available in the channel is divided between several users and each user is allotted a time slice
When a program is being executed in an 8085 microprocessor, its program counter contains
Answer: D. The memory address of the instruction that is to be executed next
The sum S of A and B in a half adder can be implemented by using K NAND gates. The value of K is
Answer: B. 4
Explanation: A half adder's sum output is represented by the XOR operation ($A \oplus B$). Implementing a 2-input XOR gate using exclusively 2-input NAND gates requires a minimum of 4 NAND gates.
VSWR of a transmission line is always
Answer: B. Greater than unity
Explanation: $1 \le VSWR < \infty$. So, the transmission line is having VSWR always greater than unity.
Wave guide acts like a
Answer: A. High pass filter
Explanation: A waveguide allows frequencies above a certain cutoff frequency to propagate while attenuating lower frequencies, thus acting as a high pass filter.
The wave length of a 100 MHz electromagnetic wave propagating through a perfect non magnetic dielectric with relative permittivity $\epsilon_r=9$ is
Answer: C. 100 cms
Explanation: $\lambda = \frac{v}{f} = \frac{c}{\sqrt{\epsilon_r} \cdot f} = \frac{3\times 10^8 / \sqrt{9}}{100\times 10^6} = \frac{1\times 10^8}{10^8} = 1\text{ m} = 100\text{ cm}$.
TEM mode exists in
Answer: C. A co-axial cable
Explanation: Transverse Electromagnetic (TEM) mode requires at least two conductors to exist. Coaxial cables have two conductors (inner and outer), whereas standard rectangular or circular waveguides are single-conductor hollow structures and do not support TEM mode.
The signal received from a circularly polarized signal by an antenna with linear polarization compared to the signal received by same circularly polarized antenna will be
Answer: C. 3 dB less
Explanation: When receiving a circularly polarized wave with a linearly polarized antenna, there is a polarization mismatch loss of 0.5, or 3 dB.
When the antenna diameter is doubled, the gain of the antenna
Answer: D. Increases by 6 dB
Explanation: Antenna gain is proportional to the square of the diameter ($G \propto D²$). Doubling the diameter increases the gain by a factor of 4. In decibels, $10\log_{10}(4) \approx 6\text{ dB}$.
Intrinsic impedance of free space is given as
Answer: C. 377 $\Omega$
Explanation: Intrinsic impedance of free space $\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 120\pi \approx 377~\Omega$.
Mark the incorrect relation
Answer: D. $B=\mu D$
Explanation: $B=\mu D$ is incorrect. The correct constitutive relations are $D=\epsilon E$ and $B=\mu H$.
If the PRF is 1200 and the pulse width is 1.5 $\mu$s, the duty cycle will be
Answer: C. 0.18 percent
Explanation: Pulse Repetition Time $T = \frac{1}{PRF} = \frac{1}{1200} = 8.33\times 10^{-4}$ s.\nDuty cycle $= \frac{\text{Pulse Width}}{T}\times 100 = \frac{1.5\times 10^{-6}}{8.33\times 10^{-4}}\times 100 = 0.18\%$.
When VSWR is 3, reflection coefficient is
Answer: A. 1/2
Explanation: $VSWR = \frac{1+|\Gamma|}{1-|\Gamma|} \Rightarrow 3 = \frac{1+|\Gamma|}{1-|\Gamma|} \Rightarrow |\Gamma| = \frac{1}{2}$
Which transmission line is ideal for handling high power?
Answer: D. Rectangular waveguide
Explanation: Rectangular waveguides have large cross-sectional areas and no center conductor, making them ideal for handling high power with minimal dielectric breakdown or heating compared to coaxial lines.
Noise figure is defined as
Answer: A. $F=\frac{S_i/N_i}{S_0/N_0}$
Explanation: Noise figure (F) is the ratio of the signal-to-noise ratio at the input to the signal-to-noise ratio at the output.
Capture effect is a characteristic of
Answer: B. FM system
Explanation: The capture effect is a phenomenon in FM (Frequency Modulation) reception where, if two signals are present at or near the same frequency, the receiver will completely suppress the weaker signal and only output the stronger one.
In a band limited channel higher bit rate can be transmitted with
Answer: B. QPSK
Explanation: QPSK (Quadrature Phase Shift Keying) transmits 2 bits per symbol, allowing for a higher bit rate in the same bandwidth compared to BPSK (which transmits 1 bit per symbol).
In a transmission line terminated with a load equal to the characteristic impedance, the reflection coefficient is
Answer: A. zero
Explanation: Reflection coefficient $(\Gamma) = \frac{Z_L-Z_0}{Z_L+Z_0}$. If the load matches the characteristic impedance ($Z_L = Z_0$), then $(\Gamma) = \frac{0}{2Z_0} = 0$.
Pointing vector $P=E \times H$ has the unit
Answer: A. Watts/metre$^2$
Explanation: The Poynting vector represents the directional energy flux (the energy transfer per unit area per unit time) of an electromagnetic field, measured in Watts per square meter.
If 1 watt of RF power is fed to a directional coupler having 30 dB coupling, the power available at the coupled port is
Answer: D. $\frac{1}{1000} W$
Explanation: $Coupling (dB) = 10 \log_{10}(\frac{P_{in}}{P_{coupled}})$. With $P_{in} = 1$ W and Coupling = 30 dB, $30 = 10 \log_{10}(\frac{1}{P_{coupled}}) \Rightarrow 3 = \log_{10}(\frac{1}{P_{coupled}}) \Rightarrow \frac{1}{P_{coupled}} = 1000 \Rightarrow P_{coupled} = \frac{1}{1000}$ Watt.
The following demodulator scheme requires least $E_b/N_0$
Answer: A. BPSK
Explanation: Among the given digital modulation schemes, BPSK (Binary Phase Shift Keying) has the lowest probability of error for a given $E_b/N_0$, meaning it requires the least signal energy per bit for reliable reception.
The channel capacity under the Gaussian noise environment for a discrete memoryless channel with a bandwidth of 4 MHz and SNR of 31 is
Answer: A. 20 Mbps
Explanation: According to Shannon's channel capacity theorem: $C = B \log_2(1+SNR) = 4 \log_2(1+31) = 4 \log_2(32) = 4 \times 5 = 20$ Mbps.
Satellite channel can be attributed
Answer: C. Both bandwidth and power limited
Explanation: Satellite channels face strict limitations on both available frequency spectrum (bandwidth) and the solar power/battery capacity generated onboard the spacecraft (power limited).
A unit vector perpendicular to the vectors $a=2i-3j+k$ and $b=i+j-2k$, is
Answer: C. $\frac{1}{\sqrt{3}}(i+j+k)$
Explanation: $\vec{a} \times \vec{b} = \begin{vmatrix}i&j&k\\2&-3&1\\1&1&-2\end{vmatrix} = i(6-1) - j(-4-1) + k(2+3) = 5i + 5j + 5k$. Unit vector perpendicular to both is $\frac{5(i+j+k)}{\sqrt{5^2+5^2+5^2}} = \frac{1}{\sqrt{3}}(i+j+k)$.
The region of the z plane for which $|\frac{z-a}{z+a}|=1 (Re~a \ne 0)$ is
Answer: B. y-axis
Explanation: $|\frac{z-a}{z+a}| = 1 \Rightarrow |z-a| = |z+a| \Rightarrow |x+jy-a| = |x+jy+a| \Rightarrow (x-a)^2+y² = (x+a)^2+y² \Rightarrow x²+a²-2ax+y² = x²+a²+2ax+y² \Rightarrow 4ax = 0 \Rightarrow x = 0$. The equation $x = 0$ corresponds to the y-axis.
If $\alpha, \beta, \gamma$ are the roots of equations $x³+px²+qx+p=0$, Then the value of $\tan^{-1}\alpha+\tan^{-1}\beta+\tan^{-1}\gamma$ is
Answer: B. $n\pi$
Explanation: $\tan^{-1}\alpha+\tan^{-1}\beta+\tan^{-1}\gamma = \tan^{-1}(\frac{\alpha+\beta+\gamma-\alpha\beta\gamma}{1-(\alpha\beta+\beta\gamma+\gamma\alpha)})$. For $x³+px²+qx+p=0$, sum of roots $\alpha+\beta+\gamma = -p$, product of roots $\alpha\beta\gamma = -p$, and $\alpha\beta+\beta\gamma+\gamma\alpha = q$. Thus $\tan^{-1}(\frac{-p-(-p)}{1-q}) = \tan^{-1}(0) = n\pi$.
The value of the determinant $\begin{vmatrix}1&a&b+c\\1&b&c+a\\1&c&a+b\end{vmatrix}$ is
Answer: A. 0
Explanation: Applying the column operation $C_3 \rightarrow C_2+C_3$, the determinant becomes $\Delta = \begin{vmatrix}1&a&a+b+c\\1&b&a+b+c\\1&c&a+b+c\end{vmatrix} = (a+b+c)\begin{vmatrix}1&a&1\\1&b&1\\1&c&1\end{vmatrix}$. Since $C_1=C_3$, $\Delta=0$.
The value of $\int_0^1\int_0^1\frac{dx dy}{\sqrt{(1-x²)(1-y²)}}$ is
Answer: A. $\frac{\pi²}{4}$
Explanation: $I = \int_0^1\int_0^1\frac{dx dy}{\sqrt{(1-x²)(1-y²)}} = (\int_0^1\frac{dx}{\sqrt{1-x²}})(\int_0^1\frac{dy}{\sqrt{1-y²}})$. Since $\int_0^1\frac{dx}{\sqrt{1-x²}} = [\sin^{-1}x]_0^1 = \sin^{-1}(1) = \frac{\pi}{2}$. $I = \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi²}{4}$.
Solution of $(D²+16)y=\cos 4x$ is
Answer: B. $y=A\cos 4x+B\sin 4x+\frac{x}{8}\sin 4x$
Explanation: Auxiliary equation is $D²+16=0 \Rightarrow D=\pm 4j$. The Particular Integral (PI) = $\frac{1}{D²+16}\cos 4x = \frac{x}{2D}\cos 4x = \frac{x}{2}\int\cos 4x dx = \frac{x\sin 4x}{8}$. Therefore, the General solution is $y = A\cos 4x+B\sin 4x+\frac{x}{8}\sin 4x$.
Laplace transform of $t²+2t+3$ is
Answer: C. $\frac{2}{s³}+\frac{2}{s²}+\frac{3}{s}$
Explanation: Using the standard formula for Laplace transform: $L\{t^n\} = \frac{n!}{s^{n+1}}$. For $t²+2t+3$, $L\{t²+2t+3\} = \frac{2!}{s³}+\frac{2 \cdot 1!}{s²}+\frac{3}{s} = \frac{2}{s³}+\frac{2}{s²}+\frac{3}{s}$.
Equation of a straight line passing through the point (-1, 2) and making equal intercepts on the axes is
Answer: C. $x+y=1$
Explanation: Equation of line with equal intercepts 'a' is $x/a + y/a = 1 \Rightarrow x+y=a$. Since it passes through (-1, 2), substituting gives $-1+2=a \Rightarrow a=1$. Thus, the equation is $x+y=1$.
A bag contains eight white and six red marbles. The probability of drawing two marbles of same colour is
Answer: B. $\frac{8c_2}{14c_2}+\frac{6c_2}{14c_2}$
Explanation: The probability of drawing two white marbles is $\frac{8c_2}{14c_2}$. The probability of drawing two red marbles is $\frac{6c_2}{14c_2}$. They are mutually exclusive, so Total probability = $\frac{8c_2}{14c_2} + \frac{6c_2}{14c_2}$.
The Algebraic multiplicity of the matrix $A=\begin{vmatrix}0&1&0\\0&0&1\\1&-3&3\end{vmatrix}$ is
Answer: A. 1
Explanation: Determinant of $A = 0 - 1(0-1) + 0 = 1$.
The impedance of an inductive reactance varies
Answer: A. Linearly with frequency
Explanation: Inductive reactance is given by $X_L = 2\pi f L$. Since $X_L \propto f$, the impedance varies linearly with frequency.
Two resistance $R_1$ and $R_2$ give combined resistance of 4.5 ohms when in series and 1 ohm when in parallel. The resistances are
Answer: C. 1.5 ohms and 3 ohms
Explanation: $R_1+R_2=4.5$ and $\frac{R_1 R_2}{R_1+R_2}=1 \Rightarrow R_1 R_2=4.5$. Using the identity $(R_1-R_2)^2=(R_1+R_2)^2-4R_1 R_2 = (4.5)^2-4(4.5) = 20.25-18 = 2.25$. So, $R_1-R_2=1.5$. Solving $R_1+R_2=4.5$ and $R_1-R_2=1.5$ gives $R_1=3~\Omega$ and $R_2=1.5~\Omega$.
Which of the following bulbs will have the least resistance?
Answer: D. 115 V, 100 W
Explanation: $R = \frac{V²}{P}$. \n$R_1 = \frac{220^2}{60} = 806.6~\Omega$. \n$R_2 = \frac{220^2}{100} = 484~\Omega$. \n$R_3 = \frac{115^2}{60} = 220.4~\Omega$. \n$R_4 = \frac{115^2}{100} = 132.25~\Omega$. \nThe bulb with 115 V, 100 W has the least resistance.
A resistance of 5 ohm is further drawn so that its length becomes double. Its resistance now is
Answer: D. 20 ohms
Explanation: When a wire is drawn to double its length, its new length $l' = 2l$. Because the volume stays constant, its new cross-sectional area becomes $A' = A/2$. New resistance $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R$. Therefore, $R' = 4 \times 5 = 20~\Omega$.
The power rating of a 470 ohm resistor carrying a current of 40 mA should be
Answer: D. 1 W
Explanation: $P = I^2R = (40\times 10^{-3})^2 \times 470 = 0.752$ watt. To safely handle this without burning out, the next standard power rating of 1 Watt should be used.
Equivalent Resistance between X and Y is \n\n[Figure needs to be uploaded later]
Answer: A. 75 $\Omega$
Explanation: [Figure needs to be uploaded later]\nFirst the rightmost two resistors 50 $\Omega$ and 25 $\Omega$ are in series, resulting in $50+25=75~\Omega$. This makes three 75 $\Omega$ resistors in parallel. $\frac{1}{R'} = \frac{1}{75}+\frac{1}{75}+\frac{1}{75} = \frac{3}{75} \Rightarrow R' = 25~\Omega$. Finally, $R_{eq} = 50 + 25 = 75~\Omega$.
The open-circuit emf of a storage cell is 2.2 volts. The terminal voltage measured when the current is 12 A is found to be 1.98 volts. The internal resistance of the cell is
Answer: B. 0.0183 ohm
Explanation: Internal voltage drop $= E - V = 2.2 - 1.98 = 0.22$ V. Internal resistance $r = \frac{\text{Voltage Drop}}{I} = \frac{0.22}{12} = 0.0183~\Omega$.
A capacitor passes a current of 12.6 mA when supplied with 20 V ac with a frequency of 1 kHz. The capacitance of the capacitor is
Answer: A. 0.1 $\mu$F
Explanation: $X_c = \frac{1}{\omega c}$ and $V = I X_c = \frac{I}{\omega c}$. $c = \frac{I}{2\pi f V} = \frac{12.6\times 10^{-3}}{2\pi \times 1\times 10^3 \times 20} = 0.1~\mu$F.
The system response can be tested better with
Answer: B. Unit impulse input signal
Explanation: A unit impulse input signal contains all frequencies uniformly, making it theoretically the best input to find the complete frequency response (transfer function) of a linear time-invariant system.
In an ideal op-amp the output impedance is
Answer: D. zero
Explanation: An ideal operational amplifier is characterized by having infinite input impedance and zero output impedance.
What will be dB gain for an increase of power level from 13 to 26 W
Answer: D. 3
Explanation: $Gain_{dB} = 10 \log_{10}(\frac{P_{out}}{P_{in}}) = 10 \log_{10}(\frac{26}{13}) = 10 \log_{10}(2) \approx 3$ dB.
The oscillator with the best frequency stability and accuracy is
Answer: D. Crystal controlled oscillator
Explanation: Crystal oscillators utilize a piezoelectric crystal which has an exceptionally high Q-factor. This results in highly stable and accurate resonant frequencies compared to standard LC oscillators.
The desirable properties of transformer core material are
Answer: C. High permeability and low hysteresis loss
Explanation: Transformer core materials should have high permeability to easily establish magnetic flux with minimal magnetizing current, and low hysteresis loss to reduce heat dissipation generated during alternating magnetic cycles.
The quality factor of series R-L-C circuit will increase if
Answer: A. R decreases
Explanation: The quality factor $Q$ for a series R-L-C circuit is inversely proportional to resistance ($Q = \frac{1}{R}\sqrt{\frac{L}{C}}$). Thus, decreasing R increases the quality factor.