Question #43751

A photocell is connected in series with a load resistor $R_L = 200 \ \text{k}\Omega$ and an ideal DC voltage source of EMF $E = 15 \ \text{V}$. The potential difference across the photocell, $V_p$, is measured by finding the null point on a potentiometer wire of length $L = 100 \ \text{cm}$. The potentiometer wire is connected across a separate ideal standard cell of $10 \ \text{V}$. In the operating region, the photocell is near saturation, and its non-ideal current-voltage characteristic is given by $I = I_0 + \frac{V_p}{r_d}$, where $I_0$ is the ideal saturation current proportional to the incident light intensity, and $r_d = 200 \ \text{k}\Omega$ is the dynamic resistance of the photocell. Initially, the photocell is illuminated such that $I_0 = 25 \ \mu\text{A}$, and the potentiometer null point is obtained at a distance $x_1$ from the high-potential end. The light intensity is then increased such that $I_0$ becomes $45 \ \mu\text{A}$, and the new null point is at $x_2$. Calculate the magnitude of the shift in the null point, $|\Delta x| = |x_2 - x_1|$.