Solved Anna University Question Paper with Answer - 1 (PART A)

SOLVED ANNA UNIVERSITY QUESTION PAPER

Question Paper Code: 40036

B.E. DEGREE EXAMINATION, APRIL/MAY 2018

Second Semester

Electrical and Electronics Engineering

EE 8251-CIRCUIT THEORY

(Common to Electronics and Instrumentation Engineering and Instrumentation and Control Engineering)

(Regulation 2017)

Time: Three Hours Maximun: 100 Marks

Answer ALL questions

PART-A (10 x 2 = 20 Marks)

1. State Kirchoff's current law.

The algebraic sum of current in a node is zero

-I₁ - I₂ + I₃+I₄= 0.

2. Estimate the resultant resistance produced by the parallel connection two resistors of 10Ω and 30 Ω.

1/R = 1/R₁ + 1/ R₂

1/R = 1/10 + 1/30

1/R = 0.1 + 0.033

R = 7.5 Ω

3. Draw the circuit of a pratical voltage source and its equivalent current source.

Practical voltage source

Current source

4. Let a network has Thevenin's equivalent circuit with source of 50V and impedance of 50 j30 Ω. Find optimum value of load to derive maximum power from the network.

P_max = v²/4 R_L

5. Define time constant and write the time constant of a series RC circuit.

The time constant for the RC circuit is the time during which the current falls to 36.8% of initial current.

6. Let a RL circuit has 50 Ω and 1 mH elements and free of source but, the inductor has initial current of 1 mA at time t=0s. Find the voltage across the resistor at time t=∞.

R = 50 Ω, L = 1×10^-3, io=1×10^-3 A

R_i + L di/dt = 0

50i + 1×10^-3 di/dt = 0

 Take laplace

Voltage across R = 50 × 0.00002

Voltage across = 1 mV.

7. Draw the phasor diagram of voltages derived from a 3-phase source.

8. In a reactive circuit, the current leads the voltage by angle 45°. Find whether the resultant reactive is either inductive capacitive and power factor.

9.Comment on the impedance and phase angle between voltage and current at resonance.

If X_L = X_C, the impedance is minimum and equal to resistance at resonance the applied voltage and current are in phase.

10. If circuit resonates at 1 MHz and produces - 3dB hand width of 100kHz

then, find the quality factor of the circuit.

f₀ = 1×10⁶

f₂ - f₁ = 100×10³

Q = f₀ /f₂ – f₁

Q = 1×10⁶  / 100×10³

Q = 10