A) 10 mA
B) 40 Ma
C) 80 mA
D) 160 mA
Correct Answer: C
Solution :
In space charge limited region, the plate current is given by Child?s law \[{{i}_{p}}\,\,=\,\,KV_{p}^{3/2}\] \[Thus,\,\,\frac{{{i}_{{{p}_{2}}}}}{{{i}_{{{p}_{1}}}}}\,\,=\,\,{{\left( \frac{{{V}_{{{p}_{2}}}}}{{{V}_{{{p}_{1}}}}} \right)}^{3/2}}\,\,=\,\,{{\left( \frac{600}{150} \right)}^{3/2}}\,=\,\,{{(4)}^{3/2}}\,=\,\,8\] \[or\,\,{{i}_{{{p}_{2}}}}-{{i}_{{{p}_{1}}}}\times \,\,8=10\times 8\,\,mA\,\,=\,\,80\,\,mA\]
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