Charge = Current/time
Q = It
t = Q/I = 15/20 = 0.75 s
Answer:
a) If the thermometer is placed well below the condenser it will record a higher temperature.
b) If the thermometer is placed well above the condenser it will record a lower temperature.
Explanation:
A) ) If the thermometer is placed below the opening to the condenser, a higher temperature will be recorded than if it is placed close to the opening since the bulb is contact with both the vapuor and liquid that is entering the condenser.
B) If the thermometer is placed above the condenser, a lower temperature will be recorded than if it is placed close to the opening since the bulb is not in full contact with both the vapuor and liquid that is entering the condenser.
I = V/Z
V = voltage, I = current, Z = impedance
First let's find the total impedance of the circuit.
The impedance of the resistor is:
= R
R = resistance
Given values:
R = 1200Ω
Plug in:
= 1200Ω
The impedance of the inductor is:
= j2πfL
f = source frequency, L = inductance
Given values:
f = 59Hz, L = 2.4H
Plug in:
= j2π(59)(2.4) = j889.7Ω
Add up the individual impedances to get the Z, and convert Z to polar form:
Z = +
Z = 1200 + j889.7
Z = 1494∠36.55°Ω
I = V/Z
Given values:
V = 170∠0°V (assume 0 initial phase)
Z = 1494∠36.55°Ω
I = 170∠0°/1494∠36.55°Ω
I = 0.1138∠-36.55°A
Round the magnitude of I to 2 significant figures and now you have your maximum current:
I = 0.11A
It transforms it to mechanical
