Answer:
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answer :
20.16 v
Explanation:
The reading of the voltmeter at the instant the switch returns to position a
L = 5H
i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo
= 1/5 ∫ 3*10^-3 d(t) + 0 = 0.6 * 10^-3 t
iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA
Rm ( resistance ) = 21 * 1000 = 21 kΩ
The reading of the voltmeter ( V )
V = IR
= 0.96 mA * 21 k Ω = 20.16 v
Solution:
Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg
Temperature = 200 °C
And 25% of the volume by liquid water is steam.
State 1
(taking the value of
and
at 200°C )
Now quality of vapor
Internal energy at state 1 can be found out by
= 856.54 kJ/kg
After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor
Tank is rigid, so volume of tank is constant.
Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature
= 369.11° C
Internal energy at state 2
Now power rating of the resistor
= 1.51 kW