  Question:
Published on: 22 June, 2022

The galvanometer shown in the circuit has a resistance of 5 Ω. Find the current through the galvanometer using Thevenin’s theorem. 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 $$V_{o.c}=V_y-V_x=6-5.714=0.286\ V$$ Fig. 22(b) Fig. (c) Fig. 22(d)

Here the galvanometer resistance $$R=5\ \mathrm{\Omega}$$

The current through the galvanometer is $$I_L=\frac{V_{o.c}}{R_{th}+R}=\frac{0.286}{12.86+5}=0.016\ Amp$$.

Random questions