To convert galvanometer into ammeter, shunt of $0.01\,\Omega $ is used. Resistance of galvanometer coil is $50\,\Omega $ and its maximum deflection current is $20\ mA$ . Range of ammeter is
A$0-1\ A$
B$0-10\ A$
C$0-100\ A$
D$0-1000\ A$
Medium
Download our app for free and get started
C$0-100\ A$
c $\mathrm{i}_{\mathrm{S}} \mathrm{R}_{\mathrm{S}}=\mathrm{i}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}$
$\frac{20 \times 50}{1000}=i_{S} \times 0.01$
$\mathrm{i}_{\mathrm{S}}=100 \mathrm{\,A}$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A proton (mass $ = 1.67 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C)$ enters perpendicular to a magnetic field of intensity $2$ $weber/{m^2}$ with a velocity $3.4 \times {10^7}\,m/\sec $. The acceleration of the proton should be
At a distance of $10\, cm $ from a long straight wire carrying current, the magnetic field is $0.04\, T$. At the distance of $40\, cm$, the magnetic field will be....$T$
The magnetic field due to a current carrying circular loop of radius $3\, cm$ at a point on the axis at a distance of $4\, cm$ from the centre is $54\, \mu T$. What will be its value at the centre of the loop.......$\mu T$
An insulating thin rod of length $l$ has a linear charge density $\rho \left( x \right) = {\rho _0}\,\frac{x}{l}$ on it. The rod is rotated about an axis passing through the origin $(x = 0)$ and perpendicular to the rod. If the rod makes $n$ rotations per second, then the time averaged magnetic moment of the rod is
A long solenoid of radius $1\,mm$ has $100 $turns per $mm$. If $1\,A$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is: