Question #252603

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


1
Expert's answer
2021-10-18T11:04:59-0400


resistance of galvanometer is 17 Ω17\ \Omega


For loop ABDA:

55IAB+17IG21IBD=055I_{AB}+17I_G-21I_{BD}=0


For loop BCDB:

25(IABIG)45(IBD+IG)17IG=025(I_{AB}-I_G)-45(I_{BD}+I_G)-17I_G=0


For loop ADCEA:

21IBD+45(IBD+IG)=2421I_{BD}+45(I_{BD}+I_G)=24


So, we have:

55IAB+17IG21IBD=055I_{AB}+17I_G-21I_{BD}=0

25IAB87IG45IBD=025I_{AB}-87I_G-45I_{BD}=0

66IBD+45IG=24    22IBD+15IG=866I_{BD}+45I_G=24\implies 22I_{BD}+15I_G=8


IBD=0.360.68IGI_{BD}=0.36-0.68I_G


55IAB+31.28IG=7.5655I_{AB}+31.28I_G=7.56

25IAB56.4IG=16.225I_{AB}-56.4I_G=16.2


IAB=0.65+2.26IGI_{AB}=0.65+2.26I_G

35.75+124.3IG+31.28IG=7.5635.75+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:

1R=117+145+25+155+21==5320+1292+119090440=780290440\frac{1}{R}=\frac{1}{17}+\frac{1}{45+25}+\frac{1}{55+21}=\frac{}{}=\frac{5320+1292+1190}{90440}=\frac{7802}{90440}


R=904407802=11.6 ΩR=\frac{90440}{7802}=11.6\ \Omega


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