question_answer

Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in ammeter a shunt is used?

A straight horizontal conducting rod of length 0.45 m and mass 60g is suspended by two vertical wires at its ends. A current of 5 A is set up in the rod through the wires.

(i) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

(ii) What will be the total tension in the wires, if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires) \[(g=9.8\,m{{s}^{-2}}).\]

Answer:

(i) To convert a galvanometer into voltmeter, its resistance needs to be increased, because a voltmeter is connected in parallel with the circuit element which voltage is to be measured. Due to its high resistance, it draws a very small current. Therefore, the potential difference across the element remains practically unaffected. A high resistance is connected in series with the galvanometer, then the value of R is given by \[R=\frac{V}{{{I}_{g}}}-G\] To convert a galvanometer into ammeter, its resistance needs to be lowered, so that maximum current can pass through it and it can give exact reading. Thus, a shunt (low resistance) is connected in parallel with the galvanometer. \[S=\frac{{{I}_{g}}}{I-{{I}_{g}}}\times G\] Or             Length of the rod, \[I=0.45\,m\] Mass suspended by the wires, \[m=60g=60\times {{10}^{-\,3}}\,kg.\] Acceleration due to gravity, \[g=98\,m/{{s}^{2}}.\] Current in the rod flowing through the wire, \[I=5A.\] (i) \[\therefore \] \[{{F}_{m}}=IIB\] for tension becomes zero in supporting wires, then \[\Rightarrow \]   \[IBI=mg\] \[\therefore \]      \[B=\frac{mg}{II}=\frac{60\times {{10}^{-\,3}}\times 9.8}{5\times 0.45}\] \[=026T\] The magnetic field should be such that Fleming?s left hand rule given an upward magnetic force. (ii) If the direction of current is reversed, then the force due to magnetic field and the weight of the wires acts in a vertically downward direction. \[\therefore \] Total tension in the wire \[=BIl+mg\] Putting all values, we get \[T=1.173\,N\approx 1.2\,N\]