In the circuit shown in the figure,

Options

A. 0.5 A
B. 0.2 A
C. 0.041666666666667
D. 0.083333333333333

Answer

A

Explanation

Above circuit is satisfied, the Wheatstone bridge condition, so no current is flow in the branch BC and it behaves like an open circuit.
So, circuit becomes as shown below
Now, in the above circuit
`R_(ABD) = 5+10 = 15 Omega`
`R_(ACD) = 10+20 = 30Omega`
Resistance, `R_(ABD)` and `R_(ACD)` in parallel, so
`R_(net) = (15 xx 30)/(30+15)=450/45=10Omega`
`therefore i=E/(R_(net) =5/10 = 0.5`A