Question

A galvanometer having a resistance of 80 ohm is shunted by a wire of resistance 20 ohm. If the total current is ten ampere, the part of it passing through the shunt is.

Answer 1

Hint: Galvanometer can be converted into ammeter and can be converted to voltmeter. It depends on the type of connection we give to external resistance. If we connect small resistance in parallel to the galvanometer then it would become an ammeter and if we connect big external resistance to the galvanometer in series then it would become voltmeter.

Formula used:
$\eqalign{
  & V = IR \cr
  & {I_1} = \dfrac{{{R_2}}}{{{R_1} + {R_2}}}I \cr} $

Complete step by step answer:
In case of parallel connection of resistors which means that the voltage across the two resistors will be same and current passing through the resistors vary.
We have
$V = IR$
Where V is the voltage and I is the current and R is the resistor.
If it is in a parallel connection then voltage will be the same and current is inversely proportional to voltage.
If we assume shunt resistance as ${R_1}$i.e 20 ohms here and galvanometer resistance which is 80 ohms as ${R_2}$ and the current passing through shunt be ${I_1}$ and the current passing through the galvanometer be ${I_2}$ and let total current be $I$which is given as 10 amperes
Now since current passing is inversely proportional to resistance we have current passing through shunt as
${I_1} = \dfrac{{{R_2}}}{{{R_1} + {R_2}}}I = \dfrac{{80}}{{100}} \times 10 = 8A$
So out 10 amperes 8 amperes pass through shunt and 2A pass through galvanometer
Hence the answer would be 8 amperes. It is evident that if resistance is less more current is passing and if resistance is more than less current is passing through that. Resistance is nothing but the obstruction or hindrance to the flow of current.

Note:
In case of conversion of galvanometer to ammeter we will attach a shunt resistance in parallel to galvanometer to get the small resistance ammeter and we will connect that ammeter in series with the circuit where we should measure the current where as if we connect external high resistance in series with galvanometer to make high resistance voltmeter and we connect that voltmeter in parallel to the circuit where we want to measure the voltage.