Answer to Question #252603 in Electric Circuits for Glenn

Question

calculate the current through the galvanometer and obtain the effective resistance of the network

Answer 1

resistance of galvanometer is $17\ \Omega$

For loop ABDA:

$55I_{AB}+17I_G-21I_{BD}=0$

For loop BCDB:

$25(I_{AB}-I_G)-45(I_{BD}+I_G)-17I_G=0$

For loop ADCEA:

$21I_{BD}+45(I_{BD}+I_G)=24$

So, we have:

$55I_{AB}+17I_G-21I_{BD}=0$

$25I_{AB}-87I_G-45I_{BD}=0$

$66I_{BD}+45I_G=24\implies 22I_{BD}+15I_G=8$

$I_{BD}=0.36-0.68I_G$

$55I_{AB}+31.28I_G=7.56$

$25I_{AB}-56.4I_G=16.2$

$I_{AB}=0.65+2.26I_G$

$35.75+124.3I_G+31.28I_G=7.56$

the current through the galvanometer:

$I_G=(7.56-35.75)/(124.3+31.28)=-0.18\$A

the effective resistance of the network:

$\frac{1}{R}=\frac{1}{17}+\frac{1}{45+25}+\frac{1}{55+21}=\frac{}{}=\frac{5320+1292+1190}{90440}=\frac{7802}{90440}$

$R=\frac{90440}{7802}=11.6\ \Omega$