current
Could someone please help me to get a better understanding as to how to calculate the current flowing through the resistor and the voltage drop across the silicon diode with the assumption that the silicon diode has a forward-biased junction potential of 0.7V
My favorite educator, Bill Beaty, often rants at the many misconceptions that all too many people have been infected with. One of the many common misconceptions involves batteries.
"Frequently-Asked Electricity Questions": "THE LIQUID BETWEEN A BATTERY'S PLATES IS A GOOD CONDUCTOR. SO WHY DOESN'T IT SHORT OUT THE BATTERY?"
"Why is electricity so hard to understand?" "...mistaken belief that no charge flows through batteries. ... This leads to the traditional incorrect flashlight-current explanation (current comes out of battery, flows...etc.) It also leads to the misconception that batteries SUPPLY CHARGE, and have a storage place for "used" charge. This might make sense if we believe that there's no path for charge through the battery. But it's wrong, because there is a path, a path provided by flowing charged atoms. Charge must flow around and around a circuit, passing THROUGH the battery over and over."
"But how SHOULD we teach kids about 'electricity'?" "A battery is a chemically-fueled charge pump. Like any other pump, a battery takes charges in through one connection and spits them out through the other. A battery is not a source of the "stuff" being pumped. When a battery runs down, it's because its chemical fuel is exhausted, not because any charges have been lost. ... When you "recharge" a battery, you are pumping charges through it backwards, which reverses the chemical reactions and converts the waste products back again into chemical fuel."
'Which way does the "electricity" really flow?' "When you connect a lightbulb to a battery, you form a complete circuit, and the path of the flowing charge is through the inside of the battery, as well as through the light bulb filament. Battery electrolyte is very conductive."
The big problem is that you didn't draw the schematic properly. A good schematic always makes the things clear:
simulate this circuit – Schematic created using CircuitLab
The current is \$I = \frac{U}{I}\$
I0 = I1+I2
I0 = (5V-2.065V)/270 = 10.87037037mA
I1 = (2.065V - 1.863V)/100 = 2.02mA
I2 = I0 - I1 = 8.85037037mA
Best Answer
This is actually rather simple, given your assumption. We can apply kirchoff's voltage law, which states that the sum of the voltages around a loop equals zero. $$V_{in} - V_R - V_D = 0$$ With \$V_{R}\$ The voltage over the resistor, \$V_{D}\$ voltage over the diode and \$V_{in}\$ the voltage you apply to the circuit. Solving for \$V_{R}\$: $$V_R = V_{in} - V_D = V_{in} - 0.7$$ Simply fill this in in Ohm's law: $$I = V_R / R = (V_{in} - 0.7)/R$$ Edit : This formula is only valid when \$V_{in} > 0.7V\$ as Warren said in the comments. A voltage lower than that would not induce any negative current. (Assuming an ideal diode)