Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is 
Explanation:
Generally for an n-type semiconductor the current density is mathematically represented as
Here 
is mathematically represented as
=> 
=> 
So
From the diagram 
=> 
So
So from
substitute
for q and 
and from the diagram 
So
Answer:
Explanation:
Given that,
AC frequency of 2.3KHz
f=2.3×10³Hz
Vrms produce is
Vrms=1.5V
Current rms
Irms= 31mA
The capacitor is reconnected to a generator of frequency
f=4.8KHz =4800Hz
The current rms becomes
Irms= 85mA
Vrms=?
Solution
First genrator
The capacitive reactance is given as
Xc=Vrms/Irms
Xc=1.5/31×10^-3
Xc=48.39 ohms
Now, to know the capacitance of the capacitor
Xc=1/2πfC
Then,
C=1/2πfXc
So,
C=1/2×π×2300×48.39
C=1.43×10^-6C
C=1.43μF
Note: the capacitance of the capacitor did not change,
Now for generator two.
The reactance are given as
Xc=1/2πfC
Xc=1/2×π×4800×1.43×10^-6
Xc=23.19ohms
Then,
Vrms2=Irms2 ×Xc
Vrms2=85×10^-3×23.19ohms
Vrms2=1.97V
Vrms2=1.97Volts
The answer would be B. Magnetic fields are invisible they cannot be directly observed
Answer:
From the negative to the positive cable.
Explanation:
The electrons have negative charge, which means that the negative terminal of the battery will suply the electrons, thus they are present in excess on the negative cable and will jump from it to the positive cable. This current direction is called real current.
