basiccurrent
"Calculate V (across the current source) and the current flowing down through R1."
simulate this circuit – Schematic created using CircuitLab
Does the voltage source also produce current, so that the current flowing to the parallel resistors is the sum of 40mA and the current produced by the voltage source? (For knowing which value to input into current division formula)
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
Just another method. The key point is to judge if the diode is on.
Assume the LED is on, then the voltage drop on the LED and R2 should be 2V or so, then
$$ V_{D1} = V_{R2} = 2V $$
Then the current on \$R_{2}\$ should be
$$ I_{R2}= \frac{2V}{200\Omega}=10mA $$
Then the voltage drop on \$R_{1}\$ should at least
$$ V_{R1} = (I_{R2} + I_{D1}) \times R_{1} >10mA \times 1k\Omega=10V $$
Apparently, this voltage is greater than the source voltage, so the diode can NOT be on. So we think it's open circuit. Then it's a voltage source in series with two resistor now. And the math will be easy.
Best Answer
If two components are in series, then the current through them is equal. That means the current through the voltage source is equal to the current through the current source.