Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
Voltage across the electric bulb, E = 12 volts
Let E0 be the peak value of voltage.
We know that heat produced by passing an alternating current ( i ) through a resistor is equal to heat produced by passing a constant current `(i_{rms})`through the same resistor. If R is the resistance of the electric bulb and T is the temperature, then
`i^2RT = i^2_rms^{RT}`
`⇒ E^2/R^2 = E_{rms}^2/R^2`
`⇒ E^2 = E_0^2/2 (therefore E^2_rms = E_0^2)`
`⇒ E_0^2 = 2E^2`
`⇒ E_0^2 = 2xx(12)^2`
⇒ E02 = 2 × 144
`⇒ E_0 = sqrt(2 xx 144)`
`⇒ E_0 = sqrt(2) xx 12`
`⇒ E_0 = 1.4142 xx 12`
⇒ E0 = 16.9704
= 16.97 ≈ 17 V
Thus , peak value of voltage is 17 V.
APPEARS IN
संबंधित प्रश्न
A device X is connected across an ac source of voltage V = V0 sin ωt. The current through X is given as
`I = I_0 sin (omega t + pi/2 )`
1) Identify the device X and write the expression for its reactance.
2) Draw graphs showing the variation of voltage and current with time over one cycle of ac, for X.
3) How does the reactance of the device X vary with the frequency of the ac? Show this variation graphically.
4) Draw the phasor diagram for the device X.
An alternating current is given by i = i1 cos ωt + i2 sin ωt. The rms current is given by
A constant current of 2.8 A exists in a resistor. The rms current is
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 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 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.
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.
The peak voltage of an ac supply is 300 V. What is the rms voltage?
A circuit containing a 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit?
[‘Average’ implies ‘averaged over one cycle’.]
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?
Phase diffn between voltage and current in a capacitor in A.C Circuit is.
In a transformer Np = 500, Ns = 5000. Input voltage is 20 volt and frequency is 50 HZ. Then in the output, we have,
RMS value of an alternating current flowing in a circuit is 5A. Calculate its peak value.
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?
