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Find the current (in A) flowing through the resistance $R_{1}$ of the circuit shown in fig, If the resistances are equal to $R_{1} = 10 Ω$ , $R_{2} = 20 Ω$ , $R_{3} = 30 Ω$ and the potential of points $1, 2 & 3$ are equal to $V_{1} = 10 V, V_{2} = 6 V$ and $V_{3} = 5 V$

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answer is 0.2.

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Detailed Solution

$V_{1} - V_{2} = I R_{1} + I_{1} R_{2}$

$V_{1} - V_{3} = I R_{1} + (I - I_{1}) R_{3}$

$5 = 40 I - 30 I_{1} \to (2)$

After solving we get $I = 0.2 A$

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