A)
B)
C)
D)
Correct Answer: D
Solution :
\[20\times 50\times {{10}^{6}}={{10}^{3}}Amp.\] \[{{V}_{1}}=\frac{2}{{{10}^{-3}}}=100+{{R}_{1}}\] \[1900={{R}_{1}}\] \[{{V}_{2}}=\frac{10}{{{10}^{-3}}}=(2000+{{R}_{2}})\] \[{{R}_{2}}=8000\] \[{{V}_{3}}=\frac{20}{{{10}^{-3}}}=10\times {{10}^{3}}+{{R}_{3}}\]\[10\times {{10}^{3}}={{R}_{3}}\]You need to login to perform this action.
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