Question

If in a diode valve, the plate potential becomes double, then plate current will be:

• A. $2\sqrt{2}$ times
• B. 8 times
• C. $3\sqrt{3}$ times
• D. 2 times

Answer 1

Answer: (A)

$2\sqrt{2}$ times

Answer 2

The Child's equation is $I_{P} \propto V_{p}^{3 / 2}$ $\therefore I_{p}'=\left(\frac{V_{p}'}{V_{p}}\right)^{3 / 2} I_{p}$ or $I_{p} =(2)^{3 / 2} I_{p}$ $=2 \sqrt{2} I_{p}$