The galvanometer shown in the circuit has a resistance of 5 Ω. Find the current through the galvanometer using Thevenin’s theorem.
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Fig.22(a)
Equivalent resistance of the circuit \(R_{th}=\frac{10\times15}{10+15}+\frac{12\times16}{12+16}=6+6.86=12.86\ \mathrm{\Omega}\)
Total current follow through the circuit \(I=\frac{10}{12.86}=0.78\ Amp\).
The current passing through the branch resistances \(12\ \mathrm{\Omega}\) and \(16\ \mathrm{\Omega}\) is \(I_1=\frac{10}{28}=0.357\ Amp\).
The current passing through the branch resistances \(10\ \mathrm{\Omega}\) and \(15\ \mathrm{\Omega}\) is \(I_2=\frac{10}{25}=0.4\ Amp\).
\(V_x=10-12\times0.36=10-4.286=5.714\ Amp\).
And \(V_y=10-0.4\times10=10-4=6\ V\)
Hence \(V_{o.c}=V_y-V_x=6-5.714=0.286\ V\)
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Fig. 22(b)
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Fig. (c)
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Fig. 22(d)
Here the galvanometer resistance \(R=5\ \mathrm{\Omega}\)
The current through the galvanometer is \(I_L=\frac{V_{o.c}}{R_{th}+R}=\frac{0.286}{12.86+5}=0.016\ Amp\).
Find the current in each branch of the network using Kirchhoff’s law
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