Can anyone help me with these questions? TIA! (Don’t actually answer please! :) )
1 answer:
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7:
</h2>
The graph of
• The I-V for Ohmic Metal wire conductor at constant temperature always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1
• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2
<h2>_____________________________________ </h2><h2>Question 8: </h2>
A diode is a device that allows current to flow in only one direction.
Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)
Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.
The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.
<h2>_____________________________________ </h2><h2>Best Regards, </h2><h2>'Borz' </h2>You might be interested in
Answer:
2.33 nC, 4.67 nC
Explanation:
when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:
(1)
Keeping in mind the relationship between charge, voltage and capacitance:
we can re-write (1) as
(2)
where:
Q1, Q2 are the charges on the two spheres
C1, C2 are the capacitances of the two spheres
The capacitance of a sphere is given by
where R is the radius of the sphere. Substituting this into (2), we find
(3)
we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:
So the eq.(3) becomes
And re-arranging it we find:
And since we know that the total charge is
we find