To determine the resistance of the ammeter when a galvanometer is shunted with a resistance, we need to calculate the parallel combination of the galvanometer resistance and the shunt resistance.
A galvanometer can be converted into an ammeter by connecting a shunt resistance (in parallel with the galvanometer). The combined resistance \(R_A\) of the ammeter is given by the formula for parallel resistances:
\(R_A = \frac{R_g \cdot R_s}{R_g + R_s}\)
where:
- \(R_g\) is the resistance of the galvanometer = \(520 \, \Omega\)
- \(R_s\) is the shunt resistance = \(20 \, \Omega\)
Substituting the values into the formula, we get:
\(R_A = \frac{520 \cdot 20}{520 + 20}\)
\(R_A = \frac{10400}{540}\)
Upon calculating, we have:
\(R_A = 19.259 \, \Omega\)
Rounding this result, we get an effective resistance of approximately \(19.3 \, \Omega\).
Thus, the resistance of the ammeter is \(19.3 \, \Omega\). This correctly matches with the given option.
Important Questions from Moving Charge and Magnetism
A square-shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop?
The magnitude of a magnetic force on a current-carrying conductor is given by:
Under the influence of a uniform magnetic field, a charged particle moves with a constant speed v in a circle of radius r. The time period of the revolution of the particle:
A square-shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop?
The magnitude of a magnetic force on a current-carrying conductor is given by:
