The four arms of a Wheatstone bridge $($Fig. $3.19)$ have the following resistances:
$AB =100 \Omega, BC =10 \Omega, CD =5 \Omega \text {, and } DA =60 \Omega \text {. }$
A galvanometer of $15 \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across $AC$.
$AB =100 \Omega, BC =10 \Omega, CD =5 \Omega \text {, and } DA =60 \Omega \text {. }$
A galvanometer of $15 \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across $AC$.
Example-(3.7)
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Considering the mesh $\text {BADB,}$ we have
$100 I_1+15 I_g-60 I_2=0$
or $ 20 I_1+3 I_g-12 I_2=0\ {[3.65(a)]}$
Considering the mesh $\text {BCDB}$, we have
$10\left(I_1-I_g\right)-15 I_g-5\left(I_2+I_g\right)=0$
$10 I_1-30 I_g-5 I_2=0$
$2 I_1-6 I_g-I_2=0 \ {[3.65(b)]}$
Considering the mesh $\text {ADCEA},$
$60 I_2+5\left(I_2+I_g\right)=10$
$65 I_2+5 I_g=10$
$13 I_2+I_g=2 \ {[3.65(c)]}$
Multiplying Eq. $(3.65b)$ by $10$
$20 I_1-60 I_g-10 I_2=0 \ {[3.65(d)]}$
From Eqs. $(3.65d)$ and $(3.65a)$ we have
$63 I_g-2 I_2=0$
$I_2=31.5 I_g \ {[3.65(e)]}$
Substituting the value of $I_2$ into Eq. $[3.65(c)],$ we get
$13\left(31.5 I_g\right)+I_g=2$
$410.5 I_g=2$
$I_g=4.87 mA .$
$100 I_1+15 I_g-60 I_2=0$
or $ 20 I_1+3 I_g-12 I_2=0\ {[3.65(a)]}$
Considering the mesh $\text {BCDB}$, we have
$10\left(I_1-I_g\right)-15 I_g-5\left(I_2+I_g\right)=0$
$10 I_1-30 I_g-5 I_2=0$
$2 I_1-6 I_g-I_2=0 \ {[3.65(b)]}$
Considering the mesh $\text {ADCEA},$
$60 I_2+5\left(I_2+I_g\right)=10$
$65 I_2+5 I_g=10$
$13 I_2+I_g=2 \ {[3.65(c)]}$
Multiplying Eq. $(3.65b)$ by $10$
$20 I_1-60 I_g-10 I_2=0 \ {[3.65(d)]}$
From Eqs. $(3.65d)$ and $(3.65a)$ we have
$63 I_g-2 I_2=0$
$I_2=31.5 I_g \ {[3.65(e)]}$
Substituting the value of $I_2$ into Eq. $[3.65(c)],$ we get
$13\left(31.5 I_g\right)+I_g=2$
$410.5 I_g=2$
$I_g=4.87 mA .$
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