current-sourceresistorsvoltage
I am trying to understand the basics of current source, but I don't know how to calculate the voltages v1 and v2. I tried to sketch the expression for current divider, but I think I didn't actually understood how to use it properly on an exercise.
simulate this circuit – Schematic created using CircuitLab
OUT is simply a label so that the simulator can assign a voltage at that point. It is not part of a charging/discharging circuit.
The basic rule of capacitor charging is that you cannot instantly change the voltage across a capacitor (unlike a resistor). The capacitor in your circuit starts off with no energy and has 0V across it. So OUT will show as 0V. On the rising edge of the input the full voltage of the pulse appears across the 100R resistor. If the step voltage is V then the initial (charging) current will be V/100 amps. As the energy stored in the capacitor increases the voltage across it will increase (Vc). This reduces the size of the current (V - Vc)/100 amps. It is this increase in capacitor voltage that produces the characteristic exponential charging curve.
It will take ONE TIME CONSTANT (C x R) to reach about 67% of the final value.
When the input pulse returns to 0V the capacitor will start to discharge through the 100R resistor.
On the right hand circuit, \$R_{L}\$ does not have a lot of latitude, because \$Q_{2}\$'s emitter will be at -0.7V, and if \$V_{CE(sat)}\$ is for example 0.3V, the voltage across \$R_{L}\$ can only go from 0 to 0.4V.
Meanwhile, the circuit on the left hand side gives a lot more latitude, depending on where the \$R2/R1\$ voltage divider puts the base (and hence the emitter voltage, 0.7V below the base).
Best Answer
You can calculate the current through the R2/R3/R4 branch using the current divider rule. Knowing the equivalent resistance of the R2/R3/R4 combination is 50 ohms,
$$I_{234} = I_1\frac{1/50}{1/50 + 1/75}$$
$$I_{234} = (25)\frac{75}{75 + 50}$$
$$I_{234} = (25)\frac{75}{125}$$
Which is the form given by your book.