question_answer

The current I drawn from the 5 V source will be

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 Wheat?s tone'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,\[l=\frac{V}{R}=\frac{5}{10}=0.5A\]