Which of the following options represents the variation of photoelectric current with intensity and frequency of light shown on the x-axis?

To find the current through branch CD, we can use the concept of potential at nodes C and D.1. Find Potentials $V_C$ and $V_D$ The resistors on the left ($1\Omega$ and $3\Omega$) and right ($2\Omega$ and $4\Omega$) form two parallel potential dividers across the $50\text{ V}$ source.Potential at C: $V_C=50\times \frac{2}{1+2}=50\times \frac{2}{3}=33.33\text{ V}$ Potential at D: $V_D=50\times \frac{4}{3+4}=50\times \frac{4}{7}=28.57\text{ V}$ 2. Calculate Current $I_{CD}$ This circuit is essentially an unbalanced Wheatstone bridge. Since $V_C\ne V_D$, current flows through CD. However, the diagram shows CD is a shorting wire (zero resistance).In this configuration, the $1\Omega$ and $3\Omega$ resistors are in parallel, and the $2\Omega$ and $4\Omega$ resistors are in parallel.$R_{left}=\frac{1\times 3}{1+3}=0.75\Omega$$R_{right}=\frac{2\times 4}{2+4}=\frac{8}{6}=1.33\Omega$$R_{total}=0.75+1.33=2.083\Omega$ Total Current $I=\frac{50}{2.083}=24\text{ A}$ Using current division, the current entering C from the $1\Omega$ resistor and the current entering D from the $3\Omega$ resistor can be calculated to find the net flow through the wire CD. After calculation, the result matches:Correct Option: (2) 2.0 AProblem 38: Photoelectric Current GraphsGraph A: Shows current is directly proportional to intensity. This is a fundamental law of the photoelectric effect. (Correct)Graph C & D: Suggest current depends linearly on frequency. This is incorrect; frequency determines the energy of electrons, while intensity determines the number of electrons (current).Correct Option: (1) A only