Answer:
Explanation:
Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.
Energy dissipates in 55Ω resistor is given by V²/R
Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.
So for 55ohms, using voltage divider rule
V=R1/(R1+R2) ×Vt
V=55/(55+140) ×70
V=19.74Volts is across the 55ohms resistor.
Then, energy loss will be
E=V²/R
E=19.74²/55
E=7.09J
7.09J of heat is dissipated by the 55ohms resistor
None I mean you have to take a drivers test but you will have to take the knowldege test twice
It makes no sense how you typed this problem out.
Answer:
b. 0.034
Explanation:
The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:
where
R is the R-value of the material
For the insulator in this problem,
R = 29
Substituting into the equation, we find the heat transfer coefficient:
