A) zero Volt
B) 2.9 Volt
C) 13.3 Volt
D) 10.04 Volt
Correct Answer: D
Solution :
\[{{R}_{b}}=\frac{{{(120)}^{2}}}{60}=240\Omega \],\[{{R}_{H}}=\frac{240}{4}=60\Omega \] \[{{V}_{1}}=120\frac{240}{243}=120,\frac{40}{41}V\] \[{{V}_{2}}=120\frac{({{R}_{b}}||{{R}_{H}})}{({{R}_{b}}||{{R}_{H}})+6}=120\frac{48}{54}=120\frac{8}{9}V\] Loss in potential \[={{V}_{1}}-{{V}_{2}}\] \[=120\left( \frac{40}{41}-\frac{8}{9} \right)\] \[=10.40\text{ }V.\]You need to login to perform this action.
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