Advertisements
Advertisements
Question
A resistor of resistance 100 Ω is connected to an AC source ε = (12 V) sin (250 π s−1)t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.
Solution
Given:
Peak voltage of AC source, E0 = 12 V
Angular frequency, ω = 250 πs−1
Resistance of resistor, R = 100 Ω
Energy dissipated as heat (H) is given by,
`H = E_{rms}/R T`
Here,
Erms = RMS value of voltage
R = Resistance of the resistor
T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms.
`H = ∫_0^t dH`
= `∫ (E_0^2 sin^2 omega t)/R dt (therefore E_{rms = E_0 sin omega t})`
= `144/100 ∫_0^{10-3} sin^2 omegat dt`
=` 1.44/2 ∫_0^{10-3} ((1-cos 2 omegat)/2)dt`
=`1.44/2 [ ∫_0^{10-3} dt + ∫_0^{10-3} cos 2 omega t dt]`
= `0.72 [ 10^-3 - {(sin 2 omegat)/(2 omega) }]_0^(10-3)]`
`= 0.72[1/1000 - 1/(1000pi)]`
`=0.72 [1/1000 - 2/(1000pi)]`
`= ((pi - 2)/(1000pi)) xx 0.72`
`= 0.0002614 = 2.61 xx 10^-4 J`
APPEARS IN
RELATED QUESTIONS
The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for two difference photosensitive materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation.
In a series LCR circuit connected to an ac source of variable frequency and voltage ν = vm sin ωt, draw a plot showing the variation of current (I) with angular frequency (ω) for two different values of resistance R1 and R2 (R1 > R2). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance.
The voltage and current in a series AC circuit are given by V = V0cos ωt and i = i0 sin ωt. What is the power dissipated in the circuit?
Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?
Can a hot-wire ammeter be used to measure a direct current of constant value? Do we have to change the graduations?
An alternating current is given by i = i1 cos ωt + i2 sin ωt. The rms current is given by
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.
The dielectric strength of air is 3.0 × 106 V/m. A parallel-plate air-capacitor has area 20 cm2 and plate separation 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.
A capacitor of capacitance 10 μF is connected to an oscillator with output voltage ε = (10 V) sin ωt. Find the peak currents in the circuit for ω = 10 s−1, 100 s−1, 500 s−1 and 1000 s−1.
Answer the following question.
A small town with a demand of 1200 kW of electric power at 220 V is situated 20 km away from an electric plant generating power at 440 V. The resistance of the two wirelines carrying power is 0.5 Ω per km. The town gets the power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.
Do the same with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?
The period of oscillation of a simple pendulum is T = `2π sqrt"L"/"g"`. The measured value of L is 20.0 cm known to have 1 mm accuracy and the time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of ls resolution. The accuracy in the determination of g is:
Phase diffn between voltage and current in a capacitor in A.C Circuit is.
When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220V. This means ______.
