In the circuit diagrams $(A, B, C$ and $D$) shown below, $R$ is a high resistance and $S$ is a resistance of the order of galvanometer resistance $G$. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as
JEE MAIN 2014, Diffcult
Download our app for free and get started
The correct circuit diagram is $D$ with galvanometer resistance $G = \frac{{RS}}{{R - S}}$
Download our appand 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.*
Similar Questions
- 1For designing a voltmeter of range $50\,V$ and an ammeter of range $10\,mA$ using a galvanometer which has a coil of resistance $54\,\Omega$ showing a full scale deflection for $1\,mA$ as in figure.View Solution
$(A)$ for voltmeter $R \approx 50\,k\Omega$
$(B)$ for ammeter $r \approx 0.2\,\Omega$
$(C)$ for ammeter $r \approx 6\,\Omega$
$(D)$ for voltmeter $R \approx 5\,k \Omega$
$(E)$ for voltmeter $R \approx 500 \Omega$
Choose the correct answer from the options given below
- 2A proton is moving along $Z$-axis in a magnetic field. The magnetic field is along $X$-axis. The proton will experience a force alongView Solution
- 3View SolutionA charged particle is released from rest in a region of uniform electric and magnetic fields which are parallel to each other. The particle will move on a
- 4The magnetic induction in air at a point $1\,cm $ away from a long wire that carries a current of $1\,A$, will beView Solution
- 5A galvenometer has a resistance of $100 \,\Omega $ and a full scale range of $50\, \mu A$ . It can be used as a voltmeter or as a high range ammeter provided a resistance is added to it. Pick the correct range and resistance combinationView Solution
- 6A circular loop of radius $0.0157\,m$ carries a current of $2.0\, amp$. The magnetic field at the centre of the loop is$({\mu _0} = 4\pi \times {10^{ - 7}}\,weber/amp - m)$View Solution
- 7If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity $v$ and an electron is projected along the lines of force, what will happen to proton and electronView Solution
- 8A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop isView Solution
- 9An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e,r_p$ and ${r_\alpha }$ respectively in a uniform magnetic field $B$. The relation between $r_e,r_p$ and $\;{r_\alpha }$ isView Solution
- 10A current $i$ is flowing in a conductor of length $l$. When it is bent in the form of a circular loop then its magnetic moment will be-View Solution