Section A
DC Circuits
Question Bank
Q1) Find the current through resistance. Using superposition theorem?
Solution:
1= 0
2=
1 + 2
Q2) Find Current I using superposition theorem?
Sol: Since two voltage sources with different magnitude in parallel which cannot be connected as in single branch two different current is not possible (if 5V than I = zero).
1 =
2 =
=
1 + 2
=
Q3) Find Vth and Rth?
Sol:
Finding Isc from circuit directly:
By KCL,
Q4) Find Rth and Vth?
Sol:
Also, clear from circuit that Vth = 1V.
By applying KVL we get,
1-3Isc=0
Isc=A
Q5) Find Rth and Isc?
Sol:
Rth=3k+2k=5k
By applying KVL we get
Therefore,
Q6) Find Vx?
Solution: For Rth
By KCL,
But,
By KVL,
Q7) i?
Find
Solution: Since, no independent source is present so,
Isc = 0
And we know that,
Since Rth cannot be zero
But
Q8) Find out the Norton’s equivalent
Solution:
Since, there is no significance of current source
A
Q9) For the circuits given below write the voltage equations:
Solution: Let current i1be in loop 1 current and i2 for loop 2
Q10) For the circuits given below write the voltage equations
Solution:
For loop 1
For loop 2
Q11) Using nodal analysis find voltage across 5resistor.
Solution:
For V1
1
For V2
2
Solving 1 and 2:
For 5 voltage =
= -50.9 + 57.27
= 6.37V
Q12) Find
Sol:
Q13) Find
Sol:
Q14) Use nodal analysis to find current in the circuit?
Sol: + +3i=4
+ -3i= -3
But i=
Substituting and solving above equation we get
V2 = 10V
i= = 2A
Q15) Find equivalent resistance R?
Sol: Equivalent resistance across ab = = 4Ω
Equivalent resistance across cd = =1 Ω
Req = Rab||Rcd +3 = + 3 = 19/5 Ω
AC Circuits
Q16) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate total impedance?
Sol: Impedance Z=
Z= = 18.27ohm
Q17) For a series RLC circuit having R=12ohms, L= 0.2H, C=60F. They are connected across 100v 50Hz supply. Calculate circuit current?
Sol: I=
Z=
Z= = 13.89ohm
I = 100/13.89 =7.2A
Q18) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate power factor?
Sol: cosφ =
Impedance Z=
Z= = 18.27ohm
Cosφ = =
φ = 56.81o lagging
Q19) A coil takes a current of 6A when connected to 24V dc supply. To obtain the same current with 50HZ ac, the voltage required was 30V. Calculate inductance and p.f of coil?
Sol: The coil will offer only resistance to dc voltage and impedance to ac voltage
R =24/6 = 4ohm
Z= 30/6 = 5ohm
XL =
= 3ohm
Cosφ = = 4/5 = 0.8 lagging
Q20) The potential difference measured across a coil is 4.5V, when it carries a dc current of 8A. The same coil when carries ac current of 8A at 25Hz, the potential difference is 24V. Find current and power when supplied by 50V,50Hz supply?
Sol: R=V/I= 4.5/8 = 0.56ohm
At 25Hz Z= V/I=24/8 =3ohm
XL =
= 2.93ohm
XL = 2fL = 2x 25x L = 2.93
L=0.0187ohm
At 50Hz
XL = 2x3 =6ohm
Z = = 5.97ohm
I= 50/5.97 = 8.37A
Power = I2R = 39.28W
Q21) A coil having inductance of 50mH an resistance 10ohmis connected in series with a 25F capacitor across a 200V ac supply. Calculate resonant frequency and current flowing at resonance?
Sol: f0= = 142.3Hz
I0 = V/R = 200/10 = 20A
Q22) A 15mH inductor is in series with a parallel combination of 80ohm resistor and 20F capacitor. If the angular frequency of the applied voltage is 1000rad/s find admittance?
Sol: XL = 2fL = 1000x15x10-3 = 15ohm
XL = 1/C = 50ohm
Impedance of parallel combination Z = 80||-j50 = 22.5-j36
Total impedance = j15+22.5-j36 = 22.5-j21
Admittance Y= 1/Z = 0.023-j0.022 siemens
Q23) A circuit connected to a 100V, 50 Hz supply takes 0.8A at a p.f of 0.3 lagging. Calculate the resistance and inductance of the circuit when connected in series and parallel?
Sol: For series Z =100/0.8 = 125ohm
Cosφ =
R = 0.3 x 125 = 37.5ohm
XL = = 119.2ohm
XL = 2fL = 2x 50x L
119.2 = 2x 50x L
L= 0.38H
For parallel:
Active component of current = 0.8 cosφ = 0.3x0.3 = 0.24A
R = 100/0.24 =416.7ohm
Quadrature component of current = 0.8 sinφ = 0.763
XL= 100/0.763 = 131.06ohm
L= 100/0.763x2x50 = 0.417H
Q24) Calculate impedance of a parallel tuned RC circuit at a frequency of 400kHz and for a BW of operation equal to 20kHz. Given R=5ohm?
Sol: BW =R/2L
L= 39H
f0 =
C = 2.6x10-9F
Z= L/RC = 3kohm
Q25) Three similar resistors are connected in star across 400V 3-phase lines. Line current is 4A. Calculate the value of each resistor.
Sol: For star connection:
IL=Iph=4A
Vph=VL/ = 400/ = 231V
Rph= 231/4= 57.75ohm
For Delta Connection:
IL=4A
Iph= IL/
=4/ ==2.30A
Zph=400/2.30=173.9ohm
Rph= 173.9/3 = 57.97ohm