Electronics and Communication Engineering - Electronic Devices and Circuits
Exercise : Electronic Devices and Circuits - Section 10
- Electronic Devices and Circuits - Section 14
- Electronic Devices and Circuits - Section 27
- Electronic Devices and Circuits - Section 26
- Electronic Devices and Circuits - Section 25
- Electronic Devices and Circuits - Section 24
- Electronic Devices and Circuits - Section 23
- Electronic Devices and Circuits - Section 22
- Electronic Devices and Circuits - Section 21
- Electronic Devices and Circuits - Section 20
- Electronic Devices and Circuits - Section 19
- Electronic Devices and Circuits - Section 18
- Electronic Devices and Circuits - Section 17
- Electronic Devices and Circuits - Section 16
- Electronic Devices and Circuits - Section 15
- Electronic Devices and Circuits - Section 1
- Electronic Devices and Circuits - Section 13
- Electronic Devices and Circuits - Section 12
- Electronic Devices and Circuits - Section 11
- Electronic Devices and Circuits - Section 10
- Electronic Devices and Circuits - Section 9
- Electronic Devices and Circuits - Section 8
- Electronic Devices and Circuits - Section 7
- Electronic Devices and Circuits - Section 6
- Electronic Devices and Circuits - Section 5
- Electronic Devices and Circuits - Section 4
- Electronic Devices and Circuits - Section 3
- Electronic Devices and Circuits - Section 2
46.
A capacitor having capacitance C is raised to voltage V. The energy stored will be
Answer: Option
Explanation:
Energy stored = 0.5 CV2
47.
If the p and T circuits in figure are equivalent, then R1, R2, R3 respectively are
Answer: Option
Explanation:
Find R1, R2, R3 by delta star conversion.
48.
In an R-L circuit, the impedance is 30 Ω at a frequency of 100 Hz. At 200 Hz, the impedance should be
Answer: Option
Explanation:
Z = (R2 + ω2L2)0.5.
49.
In the circuit, Vs = 10 cos ωt, power drawn by the 4Ω resistor is 8 watt. The power factor is
Answer: Option
Explanation:
I2m = P/R 
Im = 2
Total power drawn
= 
I2R x 2 x [4 + 2] 6
VmIm cos θ = 6
x 10 x 2 x PF = 6 P.F = 0.85 .
50.
The electric energy required to raise the temperature of a given amount of water is 1000 kWh. If heat losses are 25%, the total heating energy required is __________ kWh.
Answer: Option
Explanation:
Total heating energy required = 1000 kWh + 25% of 1000 kWh.
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