Q1-Q2 Carry 1 mark each
Q3-Q5 carry 2 marks each
Negative marking: 1/3rd
Total Questions: 05
Total marks: 8
Time: 12 min.
Q1. The field coils of a dynamo have an inductance of 8H and have a resistance of 60 ohms and are connected to 120 V dc source. The steady state current in the field coiled is
(a) 15 A
(b) 2 A
(c) 0 A
(d) None of these
Q2. A spring-controlled moving iron voltmeter draws a current of 1 mA for full scale value of 100V. If it draws a current of 0.5 mA, the meter reading is
(a) 200 V
(b) 100 V
(c) 50 V
(d) 25 V
Q3. For the circuit, values V for different values of ‘R’ are shown in the table.
The Thevenin voltage and resistance of the unknown circuit are respectively.
(a) 14 V, 4Ω
(b) 4 V, 1Ω
(c) 14V, 6 Ω
(d) 10 V, 2 Ω
Q4. For the circuit of figure, some measurements were made at the terminals a-b and given in the table below.
What is the value of I_(L ) for R_L=20 Ω?
(a) 2 A
(b) 4 A
(c) 6 A
(d) 10 A
Q5. The maximum efficiency occurs in a separately excited DC generator when the terminal voltage is 220 V and the induced emf is 240 V, the stray losses, if the armature resistance is 0.2 Ω, will be:
(a) 2000 W
(b) 4000 W
(c) 1000 W
(d) 3000 W
Sol. Steady-state field current = 120/60 = 2A
Sol. For moving iron instruments, deflection ⍺ I^2.
∴Reading1/(Reading 2 ) =(I_1^2)/(I_2^2 )
∴100/V_2 =1^2/(0.5)^2 ⇒V_2=25 V.
Sol. I_a=(E_B-V)/R=(240-220)/(0.2 )=20/0.2=100 A
Copper loss=I_a^2 R_a=(100)^2×0.2=2000 W
For maximum efficiency: copper loss=constant loss=2000 W.