A 6.0-ω and a 12-ω resistor are connected in parallel across an ideal 36-v battery. What power is dissipated by the 6.0-ω res
1 answer:
Answer:
P = 216 Watts
Explanation:
Given that,
Resistance 1,
Resistance 2,
Voltage, V = 36 V
We need to find the power dissipated by the 6 ohms resistor. In case of parallel circuit, voltage throughout the circuit is same while the current is different.
Firstly, lets find current of the 6 ohms resistor such that,
V = IR
The power dissipated by 6 ohms resistor is given by :
So, the power dissipated across 6 ohm resistor is 126 watts.
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Answer:
( a ) The specific volume by ideal gas equation = 0.02632
% Error = 20.75 %
(b) The value of specific volume From the generalized compressibility chart = 0.0142
% Error = - 34.85 %
Explanation:
Pressure = 1 M pa
Temperature = 50 °c = 323 K
Gas constant ( R ) for refrigerant = 81.49
(a). From ideal gas equation P V = m R T ---------- (1)
⇒ =
⇒ Here = Specific volume = v
⇒ v =
Put all the values in the above formula we get
⇒ v = ×81.49
⇒ v = 0.02632
This is the specific volume by ideal gas equation.
Actual value = 0.021796
Error = 0.02632 - 0.021796 = 0.004524
% Error = × 100
% Error = 20.75 %
(b). From the generalized compressibility chart the value of specific volume
= v = 0.0142
The actual value = 0.021796
Error = 0.0142 - 0.021796 =
% Error = × 100
% Error = - 34.85 %