A) 11.875
B) 26.31
C) 118.75
D) None of these
Correct Answer: C
Solution :
In given circuit three resistance \[{{R}_{2}},{{R}_{4}}\] and \[{{R}_{3}}\] are parallel. \[\frac{1}{R}=\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{4}}}+\frac{1}{{{R}_{3}}}\] \[=\frac{1}{50}+\frac{1}{50}+\frac{1}{75}\] \[=\frac{75+75+50}{50\times 75}\] \[R=\frac{50\times 75}{75+75+50}=\frac{50\times 75}{200}=\frac{75}{4}\Omega =18.75\Omega \] This resistance is in series with \[{{R}_{1}}\] \ \[{{R}_{\text{resultant}}}={{R}_{1}}+R=100+18.75=118.75\Omega \]You need to login to perform this action.
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