Question 54: Current Electricity

Question

The resistance of the galvanometer and shunt of an ammeter are $$90 \mathrm{~ohm}$$ and $$10 \mathrm{~ohm}$$ respectively, then the fraction of the main current passing through the galvanometer and the shut respectively are:

Accepted Answer

To determine the fraction of the main current passing through the galvanometer and the shunt, we can use the concept of parallel resistances, since both the galvanometer and shunt are connected in parallel in an ammeter.
Let's denote:
$$R_g$$ as the resistance of the galvanometer, which is $$90 \, \Omega$$
$$R_s$$ as the resistance of the shunt, which is $$10 \, \Omega$$
$$I_g$$ as the current passing through the galvanometer
$$I_s$$ as the current passing through the shunt
$$I$$ as the main current passing through the ammeter
According to the parallel circuit rule, the voltage across both the galvanometer and shunt will be the same. So, we have:
$$R_g \cdot I_g = R_s \cdot I_s$$
Moreover, $$I = I_g + I_s$$.
First, express $$I_s$$ in terms of $$I_g$$:
$$I_s = \frac{R_g}{R_s} \cdot I_g = \frac{90}{10} \cdot I_g = 9 \cdot I_g$$
Substitute this back into the equation for the total current:
$$I = I_g + I_s = I_g + 9 \cdot I_g = 10 \cdot I_g$$
Therefore, the fraction of the main current passing through the galvanometer is:
$$\frac{I_g}{I} = \frac{I_g}{10 \cdot I_g} = \frac{1}{10}$$
And the fraction of the main current passing through the shunt is:
$$\frac{I_s}{I} = \frac{9 \cdot I_g}{10 \cdot I_g} = \frac{9}{10}$$
So, the correct option is:
Option C
: $$\frac{1}{10}$$ and $$\frac{9}{10}$$

Answer

$$\frac{1}{90}$$ and $$\frac{1}{10}$$

Answer

$$\frac{1}{10}$$ and $$\frac{1}{90}$$

Answer

$$\frac{1}{10}$$ and $$\frac{9}{10}$$

Answer

$$\frac{9}{10}$$ and $$\frac{1}{10}$$

Current Electricity

The resistance of the galvanometer and shunt of an ammeter are $90 \mathrm{~ohm} and 10 \mathrm{~ohm}$ respectively, then the fraction of the main current passing through the galvanometer and the shut respectively are:

Choose the correct answer: