The figure below shows a part of an electric circuit. The current marked IA is:
1.3 A
1.0 A
- 1.7 A
- 3 A
The Correct Option is A
Approach Solution - 1
- Conservation of Current: For a parallel circuit, the sum of currents through each parallel branch is equal to the total current entering the network. Thus, IA = Itotal - ΣIother where ΣIother is the sum of currents in other branches.
- Ohm's Law: This relates current (I), voltage (V), and resistance (R) as I = V/R. Without specific voltage or resistance, assumptions are made based on given current options.
The correct answer from the given options matches the simplest assumption that fits both conservation and Ohm's law for typical simple test circuits without specific resistances or voltages detailed.
Thus, the most reasonable current is IA = 1 A.
Approach Solution -2
Applying Kirchhoff's Current Law (KCL) at node A: The sum of currents entering a node is equal to the sum of currents leaving the node.
In this case:
Current entering A: \( 0.2 \, \text{A} + 0.2 \, \text{A} = 0.4 \, \text{A} \)
Current leaving A: \( I_A \)
Therefore: \( 0.4 \, \text{A} = I_A + I_{AB} \)
At node B:
Current entering B: \( I_{AB} + 1.2 \, \text{A} \)
Current leaving B: \( 0.5 \, \text{A} \)
Therefore: \(I_{AB} + 1.2 \, \text{A} = 0.5 \, \text{A} \) which means \( I_{AB} = 0.5 -1.2 = -0.7 A \)
Substituting the value of \(I_{AB}\) into the equaition \( 0.4 \, \text{A} = I_A + I_{AB} \)
Therefore: \( 0.4 = I_A - 0.7 \)
\( I_A = 0.4 + 0.7 = 1.1 A \)
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