A galvanometer having a resistance of $8\ \Omega$ is shunted by a wire of resistance $2\ \Omega$. If the total

Question

A galvanometer having a resistance of $8\ \Omega$ is shunted by a wire of resistance $2\ \Omega$. If the total current is $1\ 	ext{A}$, the part of it passing through the shunt will be:

Accepted Answer

1.  **Circuit Configuration:** A galvanometer (resistance $G$) converted into an ammeter has a low resistance shunt ($S$) connected in **parallel** with it .
2.  **Current Division:** The total current $I$ splits between the galvanometer ($I_g$) and the shunt ($I_s$). Since they are in parallel, the potential difference across them is equal:
    $I_g G = I_s S$
3.  **Conservation of Charge:** The total current is the sum of the individual currents: $I = I_g + I_s \implies I_g = I - I_s$.
4.  **Substitution:** Substitute $I_g$ into the voltage equation:
    $(I - I_s) G = I_s S$
    $IG - I_s G = I_s S$
    $IG = I_s (S + G)$
    $I_s = I \left( \frac{G}{S + G} ight)$
5.  **Calculation:** Given $G = 8\ \Omega$, $S = 2\ \Omega$, and $I = 1\ 	ext{A}$:
    $I_s = 1 	imes \left( \frac{8}{2 + 8} ight) = \frac{8}{10} = 0.8\ 	ext{A}$.

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