Advertisements
Advertisements
प्रश्न
An alternating current of peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used, where i is
पर्याय
14 A
about 20 A
7 A
about 10 A
Advertisements
उत्तर
about 10 A
The rms value of an alternating current is equivalent to the constant current. So, the heating effect produced is actually measured in terms of the rms value, in case of alternating current. The constant current is, thus, equal to the rms value of alternating current, which is given by,
`Irms = Ipeal = 1/sqrt2 = 14/(A 9.9 )≈ 10A`
APPEARS IN
संबंधित प्रश्न
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.
Two alternating currents are given by `i_1 = i_0 sin wt and i_2 = i_0 sin (wt + pi/3)` Will the rms values of the currents be equal or different?
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.
An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?
The peak power consumed by a resistive coil, when connected to an AC source, is 80 W. Find the energy consumed by the coil in 100 seconds, which is many times larger than the time period of the source.
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.
A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for ω = 100 s−1, 500 s−1, 1000 s−1.
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.
The peak voltage of an ac supply is 300 V. What is the rms voltage?
The rms value of current in an ac circuit is 10 A. What is the peak current?
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:
In a transformer Np = 500, Ns = 5000. Input voltage is 20 volt and frequency is 50 HZ. Then in the output, we have,
In the Figure below, the current-voltage graphs for a conductor are given at two different temperatures, T1 and T2.
- At which temperature T1 or T2 is the resistance higher?
- Which temperature (T1 or T2) is higher?
