A galvanometer has a coil of resistance $$100\,\Omega $$ and gives a full scale deflection for $$30\,mA$$ current. If it is to work as a voltmeter of $$30\,V$$ range, the resistance required to be added will be
A.
$$900\,\Omega $$
B.
$$1800\,\Omega $$
C.
$$500\,\Omega $$
D.
$$1000\,\Omega $$
Answer :
$$900\,\Omega $$
Solution :
Required resistance to convert a galvanometer into voltmeter of $$30\,V$$ is given by
$$R = \frac{V}{{{i_g}}} - G$$
Symbols have their equal meaning
$$ = \frac{{30}}{{30 \times {{10}^{ - 3}}}} - 100 = 900\,\Omega $$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Current
Releted Question 1
The temperature coefficient of resistance of a wire is 0.00125 per $$^ \circ C$$ At $$300\,K,$$ its resistance is $$1\,ohm.$$ This resistance of the wire will be $$2\,ohm$$ at.
The electrostatic field due to a point charge depends on the distance $$r$$ as $$\frac{1}{{{r^2}}}.$$ Indicate which of the following quantities shows same dependence on $$r.$$
A.
Intensity of light from a point source.
B.
Electrostatic potential due to a point charge.
C.
Electrostatic potential at a distance r from the centre of a charged metallic sphere. Given $$r$$ < radius of the sphere.