A) 0.33 A
B) 0.5 A
C) 0.67 A
D) 0.17 A
Correct Answer: B
Solution :
The given circuit can be redrawn as which is a balanced Wheatstone?s bridge and hence no current flows in the middle resistor, so equivalent circuit would be as shown below. \[10\Omega \]and \[20\Omega \]resistances are in series \[\therefore \] \[R'=10\Omega +20\Omega =30\Omega \] Similarly, \[5\Omega \]and \[10\Omega \]are in series \[R''=15\Omega \] Now, R? and R? are in parallel \[\therefore \] \[R=\frac{15\times 30}{15+30}=10\Omega \] So, \[I=\frac{V}{R}=\frac{5}{10}=0.5A\]You need to login to perform this action.
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