Question
Determine the current flowing through the galvanometer shown in the figure below.
✓
Answer
To find $I_g$ we apply Kirchhoff's voltage law.
Loop $\text{ABDA }:$
$-5 I _1-10 I _{ g }+15 I _2=0$
$\therefore-5 I _1+15 I _2=10 I _{ g }$
$\therefore- I _1+3 I _2=2 I _{ g } \ldots \ldots \ldots \ldots . . .(1)$
Loop $\text{BCDB} :$
$-10\left( I _1- I _{ g }\right)+20\left( I _2+ I _{ g }\right)+10 I _{ g }=0$
$\therefore-10 I _1+10 I _{ g }+20 I _2+20 I _{ g }+10 I _{ g }=0$
$\therefore I _1-2 I _2=4 I _{ g } \ldots \ldots \ldots \ldots . . . . .(2)$
Adding Eqs. $(1)$ and $(2),$
we get, $I _2=6 I _{ g }....(3)$
Substituting for $Z_2$ from Eq. $(3)$ in Eq. $(2).$
$\therefore I _1=12 I _{ g }+4 I _{ g }=16 I _{ g }$
Now, $I _1+ I _2=2 A$ by the data.
$\therefore 16 I _{ g }+6 I _{ g }=2 A$
$\therefore 22 I _{ g }=2 A$
$\therefore I _{ g }=\frac{2}{22} A=\frac{1}{11} A$ from $B$ to $D$
This is the current flowing through the galvanometer.
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