Compare a series circuit powered by six 1.5-volt batteries to a series circuit powered by a single 9-volt battery. Make sure the
re are equal numbers of light bulbs in each circuit and that the batteries are all in the same orientation.
1 answer:
Answer:
Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.
Explanation:
We are asked to compare two series circuits having equal number of light bulbs.
1st circuit is powered by 6 batteries each having a voltage of 1.5V
2nd circuit is powered by a single battery having a voltage of 9V.
The six batteries in the 1st circuit can be connected together in series or in parallel.
When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means
Voltage of pack = number of batteries*voltage of each battery
Voltage of pack = 6*1.5
Voltage of pack = 9 volts
But the current remains same in the series connection since there is only path for the current to flow.
On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.
Circuit 1:
In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Circuit 2:
In this circuit, we have 1 battery which provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit just like in circuit 1.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Conclusion:
Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.
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Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
Explanation:
from the T-S diagram, we get the overall pressure ratio of the cycle is 9
Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3
P5/P6 = P7/P8 = √9 =3
get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.
At temperature T1 =300K
Specific enthalpy of air h1 = 300.19 kJ/kg
Relative pressure pr1 = 1.3860
At temperature T5 = 1200 K
Specific enthalpy h5 = 1277.79 kJ/kg
Relative pressure pr5 = 238
Calculate the relative pressure at state 2
Pr2 = (P2/P1) Pr5
Pr2 =3 x 1.3860 = 4.158
get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure pr = 4.153
The corresponding specific enthalpy h = 411.12 kJ/kg
Relative pressure pr = 4.522
The corresponding specific enthalpy h = 421.26 kJ/kg
Find the specific enthalpy of state 2 by the method of interpolation
(h2 - 411.12) / ( 421.26 - 411.12) =
(4.158 - 4.153) / (4.522 - 4.153 )
h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))
h2 - 411.12 = 0.137
h2 = 411.257kJ/kg
Calculate the relative pressure at state 6.
Pr6 = (P6/P5) Pr5
Pr6 = 1/3 x 238 = 79.33
Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure Pr = 75.29
The corresponding specific enthalpy h = 932.93 kJ/kg
Relative pressure pr = 82.05
The corresponding specific enthalpy h = 955.38 kJ/kg
Find the specific enthalpy of state 6 by the method of interpolation.
(h6 - 932.93) / ( 955.38 - 932.93) =
(79.33 - 75.29) / ( 82.05 - 75.29 )
(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )
h6 - 932.93 = 13.427
h6 = 946.357 kJ/kg
Calculate the total work input of the first and second stage compressors
(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )
= 222.134 kJ/kg
Calculate the total work output of the first and second stage turbines.
(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )
= 662.866 kJ/kg
Calculate the net work done
Wnet = (Wturb)out - (Wcomp)in
= 662.866 - 222.134
= 440.732 kJ/kg
Calculate the minimum mass flow rate of air required to generate a power output of 105 MW
W = m × Wnet
(105 x 10³) kW = m(440.732 kJ/kg)
m = (105 x 10³) / 440.732
m = 238.2 kg/s
therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
Answer:
Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3
Explanation:
The step by step and appropriate substitution is as shown in the attachment.
From number of moles = Concentration x volume
number of moles = number of particles/ Avogadro's number
Volume = mass/density, the appropriate derivation to get the number of moles of atoms
