The current in the branch CD of the circuit (in ampere) is

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Solution

Let distribution of current in the circuit is as shown,

Now,
$K V L t o B D C B - --------------- - - 2 (i_{1} - i_{2}) + 2 i_{2} = 0 \Rightarrow 2 i_{2} = i_{1} K V L t o A B C A - --------------- - \Rightarrow - 2 i_{1} - 2 i_{2} + 3 (i - i_{1}) = 0 i = 2 i_{1} K V L t o A C D A via the cell - --------------------------- - - 3 (i - i_{1}) + 30 = 0 i - i_{1} = 10 \frac{i}{2} = 10 i = 20 \Rightarrow i_{1} = 10 \Rightarrow i_{2} = 5$
The current through CD is $i - i_{1} + i_{2} = 20 - 10 + 5 = 15 A$

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