The break down voltage of a transistor with its base open is $\mathrm{BV}_{\mathrm{CEO}}$ and that with emitter open is $\mathrm{BV}_{\mathrm{CBO}^{‘}}$ then
- $\mathrm{BV}_{\mathrm{CEO}}=\mathrm{BV}_{\mathrm{CBO}}$
- $\mathrm{BV}_{\mathrm{CEO}}>\mathrm{BV}_{\mathrm{CBO}}$
- $\mathrm{BV}_{\text {CEO }}<\mathrm{BV}_{\mathrm{CBO}}$
- $\mathrm{BV}_{\mathrm{CEO}}$ is not related to $\mathrm{BV}_{\mathrm{CBO}}$