Problem 4: Current in Photoelectric Circuit
Solution using Superposition:
The total current in each resistor is the algebraic sum of the current driven by the battery and the photoelectric current.
1. Battery Contribution ($I_{batt}$):
The battery sees the resistors in a series loop (Resistance of plate + Resistor A + Resistor B).
$$ I_{batt} = \frac{\mathcal{E}}{R_0 + R + R} = \frac{\mathcal{E}}{R_0 + 2R} $$
This current flows clockwise (Left to Right through the top wire).
2. Photoelectric Contribution ($i_p$):
The total photoelectric current is derived from the photon flux and quantum efficiency $\eta$:
$$ i_p = \frac{\text{Power} \times \eta \times e}{h\nu} = \frac{(I_0 S) \eta e \lambda}{hc} $$
This current enters the plate from the vacuum and splits symmetrically to return to ground through resistors A and B.
$$ I_{p, split} = \frac{i_p}{2} = \frac{e \eta \lambda I_0 S}{2hc} $$
3. Total Currents:
For Resistor A, the battery current and photocurrent flow in the same direction.
For Resistor B, they flow in opposite directions.
$$ I_B = \frac{\mathcal{E}}{R_0 + 2R} – \frac{e\eta\lambda I_0 S}{2hc} $$