Question

In an electro-discharge machining process, the discharge voltage is $V_b$. The energy dissipated per spark across the inter-electrode gap is proportional to

• A. $V_b^{0.5}$
• B. $V_b$
• C. $V_b^{2}$
• D. $V_b^{3}$

Hint:

Spark energy in EDM scales with voltage and current, and since current increases with voltage, the overall energy varies as $V_b^2$.

Answer 1

The Correct Option is C

Electro-discharge machining (EDM) is a thermo-electric erosion process in which material is removed from a conductive workpiece using controlled sparks. Each spark creates extreme local temperatures (8,000–12,000°C) that melt and vaporize a tiny portion of the material. The intensity of each spark determines how much material is removed per pulse. This intensity is governed by the energy released during the spark, which depends on both the voltage and the current flowing through the plasma channel that momentarily bridges the gap between tool and workpiece. The spark energy $E$ during a discharge pulse of duration $t_p$ is given by the general relation: \[ E = V_b I_p t_p, \] where $V_b$ is the discharge voltage and $I_p$ is the peak discharge current. In typical EDM machines, the discharge current is not constant but increases approximately linearly with the applied voltage because a higher voltage increases ionization in the dielectric fluid, widening and strengthening the plasma channel. Thus, in many practical EDM conditions: \[ I_p \propto V_b. \] Substituting this relationship in the spark energy expression, we get: \[ E \propto V_b \times V_b = V_b^2. \] This quadratic dependence implies that even small increases in discharge voltage produce disproportionately larger spark energies and therefore greatly increase the material removal rate. Neither $V_b^{0.5}$ nor $V_b^3$ align with the experimentally observed behavior of spark energy formation. Hence the energy dissipated per spark is proportional to $V_b^2$, which corresponds to option (C).