In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively :

 

Answers (1)

Let the current flowing through the circuit as shown in the diagram
From the circuit we have,

50=20I1
 

30=10I3

Further, using Kirchhoff's loop rule we get
20 \mathrm{I}_{1}=10 \mathrm{I}_{2}+10 \mathrm{I}_{3}$
Or
$2 \mathrm{I}_{1}=\mathrm{I}_{2}+\mathrm{I}_{3}$ 

or 

$\mathrm{I}_{2}=2 \mathrm{A}$
And
\mathbf{I}=\mathbf{I}_{1}+\mathbf{I}_{2}$
Thus
\mathrm{I}=4.5 \mathrm{A}$
Now,
$\mathrm{I}_{4}=\mathrm{I}_{2}-\mathrm{I}_{3}$

\mathrm{I}_{4}=-1 \mathrm{A} \text { -Ve means direction of current is going upwards }.

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