MCQ
To convert a galvanometer into a ammeter, one needs to connect a
- ✓low resistance in parallel
- Bhigh resistance in parallel
- Clow resistance in series
- Dhigh resistance in series
✓
Answer
Correct option: A.
low resistance in parallel
a
To convert a galvanometer into an ammeter, one needs to connect a low resistance in parallel so that maximum current passes through the shunt wire and ammeter remains protected.
To convert a galvanometer into an ammeter, one needs to connect a low resistance in parallel so that maximum current passes through the shunt wire and ammeter remains protected.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The potential due to an electrostatic charge distribution is $V(r)=\frac{q e^{-\alpha e r}}{4 \pi \varepsilon_{0} r}$, where $\alpha$ is positive. The net charge within a sphere centred at the origin and of radius $1/ \alpha$ is
→An air column in a pipe, which is closed at one end, will be in resonance wtih a vibrating tuning fork of frequency $264\, Hz$ if the length of the column in $cm$ is (velocity of sound $= 330\, m/s$)
→A concave mirror of focal length $15\, cm$ forms an image having twice the linear dimensions of the object. The position of the object when the image is virtual will be........$cm$
→A capillary tube of radius $0.2\,cm$ is dipped vertically in a beaker containing liquid. If the liquid rises to a height of $5\,cm$ for which the angle of contact is $60^o$ , then surface tension of the liquid is ...... $dynes/cm$ $(d = 1\,gm/cm^3)$
→$A$ long metal bar of $30\,cm$ length is aligned along a north south line and moves eastward at a speed of $10\, ms^{-1}$. $A$ uniform magnetic field of $4.0\,T$ points vertically downwards. If the south end of the bar has a potential of $0\,V$, the induced potential at the north end of the bar is.....$V$
→Two parallel wires situated at a distance $2a$ are carrying equal currents $‘i’$ in opposite direction as shown in figure. The value of magnetic filed at a point $P$ situated at equal distances from both the wires will be
→A large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y $ from the top and the other is a circular hole of radius $ R$ at a depth $ 4y $ from the top. When the tank is completely filled with water the quantities of water flowing out per second from both the holes are the same. Then $ R$ is equal to
→A non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre
→Two sound waves with wavelengths $5.0\,\, m$ and $5 .5\,\, m$ respectively, each propagate in a gas with velocity $330\,\, m/s.$ We expect the following number of beats per second.
→An amplitude modulated wave is represented by the expression $v_{m}=5(1+0.6 \cos 6280 t) \sin \left(211 \times 10^{4} t \right)\; volts$. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively
→