Starting from one terminal of source, if current has more than one path to reach the other terminal, then those two paths are parallel. I see two such paths in your circuit.
You didn't say how you calculated \$I_x\$ and \$I_o\$. The answers you got are wrong. Try Current dividision.
EDIT:
You used 4A instead of 6A in your calculations.
If you want to find the current by dividing voltage across resistance by resistance value, you have to find the voltage \$V_o\$ first.
$$V_o = 6A\times (1\Omega || 2\Omega) = 4V$$
now,
$$I_x = V_o/1\Omega = 4A$$
$$I_o = V_o/2\Omega = 2A$$
I needed to find the Light output power (LOP) and divide it with the input power to give the EQE, as EQE = Output Power / Input Power.
This is not the usual definition of quantum efficiency. Quantum efficiency should be a ratio of numbers of quanta. For an LED this would be
$$\frac{\mathrm{Photons\ emitted}}{\mathrm{charge\ carriers\ injected}}$$
If you have the input current and output power you can calculate this as
$$\frac{P_o/h\nu}{I/e}$$
where \$P_o\$ is optical output power, \$h\nu\$ is the photon energy at the emission frequency \$\nu\$, \$I\$ is the injection current, and \$e\$ is the electron charge.
You do not need the diode forward voltage to calculate the quantum efficiency. You might need it in the future if you are asked to calculate the power efficiency or wall-plug efficiency.
How am I supposed to find the total Light output power like I have with the data in spreadsheet 1 then?
The ususal way to get the total light output of an LED is to put the LED at the input port of an integrating sphere, and measure the optical power at the output port with a large-area photodiode. Then scale the measured power by the loss factor of the integrating sphere.
Or am I calculating everything wrong and the light output power is given in another way?
It looks to me like you did not collect enough data when you were doing the experiment. You need to re-do the experiment and measure the light output for all of the input current levels you are interested in.
Realistically, there's no reason the quantum efficiency (my definition, not yours) should change dramatically at current levels below 1 mA. Maybe your next experiment will be with a laser instead of an LED, then you will see some interesting behavior changes at low currents.
Best Answer
The voltage across the resistor is
$$V_{R1} = 2\mathrm A \cdot 10 \Omega = 20V$$
with the more positive terminal on the left (since the current from the current source enters the left-most resistor terminal.
Thus, the voltage across the current source (being careful to choose the reference direction such that the current enters the positive labelled terminal per the passive sign convention) is, by KVL:
$$V_{I1} = (V_{V1} - V_{R1})\mathrm V = (14 - 20)\mathrm V = -6\mathrm V$$
The power associated with the current source is then
$$P_{I1} = V_{I1} \cdot I_{I1} = -6\mathrm V \cdot 2\mathrm A = -12\mathrm W$$
By the passive sign convention, when the power is negative, the circuit element is supplying power to the circuit thus, the current source supplies \$12 \mathrm W\$ of power to the circuit.
As a check, the sum of the powers should be zero.
The power delivered to the resistor is:
$$P_{R1} = 20 \mathrm V \cdot 2 \mathrm A = 40 \mathrm W$$
The power associated with the voltage source is
$$P_{V1} = -14 \mathrm V \cdot 2 \mathrm A = -28 \mathrm W$$
The sum of the powers is then
$$P_{I1} + P_{R1} + P_{V_1} = (-12 +40 -28) \mathrm W = 0 \mathrm W$$
as required.
Note that another approach would be to calculate the resistor and voltage source powers using the series current and write, using the fact that the sum of powers is zero:
$$P_{I1} = -(P_{R1} + P_{V1}) = -\left((2\mathrm A)^2 \cdot 10 \Omega - 14\mathrm V \cdot 2\mathrm A)\right) = -12\mathrm W$$
as before.