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Question

# For the circuit shown in the figure. Find out the current in the wire BD.

A
4 A from D to B
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B
1 A from D to B
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C
3 A from D to B
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D
2 A from D to B
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Solution

## The correct option is B 1 A from D to BStep 1: Assume some current in branches taking KCL in to account at the junction Step 2: Apply KVL in closed loops, In loop DCBD, ⇒20−5i=0 i=4 A ...(1) In loop BADB, +10 − 2(i−i1) = 0 ...(2) Using Eq.(1) and (2), 10=2(4−i1) i1=−1 A The diagram showing final distribution of current in branches The negative sign of current indicates that its direction will be in the direction opposite to that assumed initially. ∴1 A current flows from D to B Method 2 Step 1 : Lets assume potential at point B to be zero. Now the points B and D are connected by same conducting wire, hence both will have same potential i.e 0 V Step 2 : Mark the potential at other points Step 3 : Find the current through each resistance by applying ohm's law i=Potential differenceResistance ⇒iCB=20−05=4 A and iAD=10−02=5 A Step 4 : Apply KCL and obtain required currents in the branches. Thus in branch BD, 1 A current flows from D to B Why this question? Tip: In such problems assigning 'Zero' potential at the nodes is recommended to find the current in respective branches, as it will save a lot of time in competitive examinations.

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