Question

State the principle of working of a galvanometer. A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance ${R_1}$ in series with the coil. If a resistance ${R_2}$ is connected in series with it can measure upto $\dfrac{V}{2}$ volts. Find the resistance, in terms of ${R_1}$and ${R_2}$, required to be connected to convert it into a voltmeter that can read upto $2V$. Also find the resistance G of the galvanometer in terms of ${R_1}$ and ${R_2}$ .

Answer 1

We want to find resistance, required to be connected to convert it into a voltmeter that can read upto $2V$, so the maximum current carrying capacity should be according to ohm’s law, it means, we can apply ohm’s law in order to find all resistances and maximum current carrying capacity.

Complete step by step answer:
Principle of working of a Galvanometer: It works on the principle of torque acting on a current carrying coil inside a magnetic field. The torque acting on this coil is directly proportional to the magnitude of electric current flowing in it. The coil is connected to a pointer and it makes the pointer of the galvanometer calibrated and hence it shows correct deflection on a scale.

Now, the resistance of the galvanometer is G (to be found)
Let maximum current carrying capacity of the galvanometer be I.
Applying ohm’s law in first condition when galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance ${R_1}$ in series with the coil, we get,
$V = I \times \left( {G + {R_1}} \right)$ ------(1)
Applying ohm’s law in first condition when galvanometer of resistance G is converted into a voltmeter to measure upto $\dfrac{V}{2}$ volts by connecting a resistance ${R_2}$ in series with the coil, we get,
$\dfrac{V}{2} = I \times \left( {G + {R_2}} \right)$ ------(2)
Dividing equation 1 by equation 2, we get,
$2 = \dfrac{{G + {R_1}}}{{G + {R_2}}}$
On further solving, we get G as,
$2G + 2{R_2} = G + {R_1} \Rightarrow G = {R_1} - 2{R_2}$
So, we got the value of G in terms of ${R_1}$ and ${R_2}$ .
Now, let the resistance to read voltage of 2V be ${R_3}$, then,
Applying ohm’s law in this, we get,
$2V = I(G + {R_3}) \\\ 2 \times \dfrac{V}{I} = G + {R_3} \\\$
Using value of $\dfrac{V}{I}$ from equation 1 we get,
$2\left( {G + {R_1}} \right) = G + {R_3}$
Using obtained value of G in above equation, we get
$2({R_1} - 2{R_2}) = {R_1} - 2{R_2} + {R_3} \\\ {R_3} = 3{R_1} - 2{R_2} \\\$
So resistance of galvanometer G is $G = {R_1} - 2{R_2}$ and resistance to read voltage of $2V$ R3 is ${R_3} = 3{R_1} - 2{R_2}$

Note: We are applying ohm’s law in every given condition to find the desired resistance values of galvanometer at different reading of voltages and since in every condition, it is given that resistance are connected in series, we are adding all given resistances and resistance to find while applying ohm’s law every time.