A circuit has a voltage source and a resistor with its terminals marked 'a' and 'b' .?

Question

if a current of 10 amperes enters the end 'a' of the resistor. will the same curfrent of 10 amperes leave the end 'b' of the resistor too. if so or if not why? will this resistor affect the currnt passing in to the diode when connected in series with it??

Answer 1

No there will be current drop at b the formula for that is V=IR, where V is the Voltage, I is the Current (10 amp), R is the resistance, yes the total current passing thro. the resistance will be the input current for diode at end b, if connected in series in a closed circuit there will be potential difference also at b. depending on the resistance of both the resistor as well as diode.

Answer 2

Kirchhoff's Current Law says that the same 10 Amps of current entering one terminal of the resistor will exit through the other terminal.  

When placed in series with a diode, the resistor will have the effect of limiting the maximum current through the diode to 10 Amps, providing the voltage of the voltage source isn't changed.

Answer 3

you have a lot of complicated answers.

Simply, the current in a series circuit is everywhere the same. So if you measure 10 amps at one point, it is 10 amps at all the other points. Why? because the current has only one path to take.

Answer 4

Okay, this is a very tricky question because it relies on a lot of assumptions:

Q1: Will the current coming out of b be the same as coming into a? why or why not.

A1: Assuming that your voltage source is ideal (meaning that it will maintain set voltage across your resistor load) and your resistive load is the constant (will not change), then current coming into the resistor will remain the same, the I (current) will also remain constant. V= I *R

Q2: Will the resistor affect the current passing into the diode when connected in series?

A1: Here is the tricky part: It depends on how the series diode is connected to the load. There are two distinct ways diodes work: (a) connecting the cathode of the diode at the end of the resistor and the anode to the b terminal yields a much different result if you were to connect by (b) flipping the diode.

Effectively, an ideal diode in the (a) configuration does not allow current to pass from the cathode to the anode. Because of this fact, the current consumed through the resistor will always (in our ideal case) equal to zero. Since this is true for any load, the resistor DOES NOT effect any current through the diode. Consequently, any resistor value there will not change this fact.

When in the flipped configuration (b), an ideal diode will always allow current to pass through. This diode state is assumed on and a small on voltage will occur across the anode to cathode. Because of this voltage, current through the resistor is no longer 10 amperes, but will be less than that value. In the flipped configuration, the resistor DOES effect the current passing through the diode.

To explain, this occurs because the V (voltage) in Ohm's Law is no longer constant. Because of the addition of diode's on voltage, the effective voltage across the resistor will always be the source minus the on voltage. Because the new voltage will be smaller than the original, the current will also decrease. The current is directly proportional to voltage!

I = V/R

I hope you can use some of this to explain the phenomenon