1. Circuit Setup and KVL

We have a series circuit containing an ideal battery $V_0$, a resistor $R$, and a nonlinear device $D$. The current $I$ flows through all components. Let $V$ be the potential difference across the nonlinear device $D$.

Applying Kirchhoff’s Voltage Law (KVL) to the loop: $$ V_0 – IR – V = 0 $$ $$ V_0 = IR + V \quad \dots (1) $$

2. Utilizing Device Characteristics

The nonlinear device follows the relation $I = kV^2$. We can substitute this into equation (1) to solve for $V$. $$ V_0 = (kV^2)R + V $$ Rearranging into a standard quadratic equation for $V$: $$ kR V^2 + V – V_0 = 0 $$

3. Solving for Voltage and Current

Using the quadratic formula to solve for $V$: $$ V = \frac{-1 \pm \sqrt{1^2 – 4(kR)(-V_0)}}{2kR} $$ Since $V$ must be positive, we take the positive root: $$ V = \frac{\sqrt{1 + 4kRV_0} – 1}{2kR} $$ Now, we find the current $I$ using $I = kV^2$: $$ I = k \left( \frac{\sqrt{1 + 4kRV_0} – 1}{2kR} \right)^2 $$ Simplifying the expression: $$ I = k \frac{(\sqrt{1 + 4kRV_0} – 1)^2}{4k^2 R^2} $$ $$ I = \frac{(\sqrt{1 + 4kRV_0} – 1)^2}{4k R^2} $$