Find out the current flowing through resistor $R_{1}$

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Solution

Resistance $R_{1}$ & $R_{2}$ are parallel and $R_{3}$ & $R_{4}$ are in series. Replace them with requirement resistance.
apply $L_{1}$ L $i n$ GEHG$,
$5 - 400 (I_{1} + I_{2}) + 5 = 0$
$\Rightarrow I_{1} + I_{2} = 4 \frac{1}{40} \to (1)$
in $A C E F A$ ,
$- 500 I_{2} - 400 (I_{1} + I_{2}) + 50 (I - (I_{1} + I_{2})) = 0$
$\Rightarrow 19 I_{2} + 9 I_{1} - I = 0 \to (2)$
in $B C D G P$ ,
$- 500 I_{1} - 5 + 5 = 0 \Rightarrow I_{2} = 0 \Rightarrow (3)$
that given $I_{1} = \frac{1}{40} A I_{2} = - 4 O A k I = \frac{19}{40} A$
through $50 σ$ , current in $= I - (I_{1} + I_{2}) = \frac{18}{40} = 0.45 A$
$\Rightarrow$ current through each $R_{1}, R_{2} = \frac{0.45}{2} = 0.225 m A$

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