Conversion of galvanometer into ammeter and voltmeter : A galvanometer is very sensitive instrument to detect the current . It can be easily converted into ammeter and voltameter.
(i) Galvanometer to an Ammeter: Ammeter is an instrument used to measure current flowing in the electrical circuit . the ammeter must offer low resistance such that it will not change the current passing through it. So ammeter is connected in series to measure the circuit current.
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer . this low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of ammeter depends on the values of the shunt resistance.
Let I be the current passing through the circuit. When current I reaches the junction A. it divides into two components . Let `I_(g ) ` be the current passing through the galvanometer of resistance `R_(g)` through a path AGE and the remaining current `(I-I_(g))` passes along the path ACDE through shunt resistance S. the value of shunt resistance is so adjusted that current `I_(g)` produces full scale deflection in the galvanometer . the potential difference across galvanometer is same as the potential difference across shunt resistance.
`V_("galvanometer") = V_("shunt") " " rArr I_(g) R_(g) = (I - I_(g)) S `
` S = (I_(s))/((I - I_(g))) R_(g) (or ) I_(g) = (S)/(S + R_(g)) I rArr I_(g) prop I`
Since, the deflection in to galvanometer is proportional to the current passing through it .
`theta = (1)/(G) I_(g) rArr theta prop I g rArr theta prop I `
So, the deflection in the galvanometer measures the current I passing through the circuit (ammeter). Shunt resistance is connected in parallel to galvanometer. therefore, resistance of ammeter can be determined by computing the effective resistance, which is
`(l)/(R_("eff")) = (1)/(R_(g)) + (1)/(S) rArr R_("eff") = (R_(g) S)/(R_(g) + S) = R_(a) `
Since, the shunt resistance is a very low resistance and the ratio `(S)/(R_(g))` is also small. This means , `R_(g)` is also small, i.e., the resistance offered by the ammeter is small. so, when we connect ammeter in series, the ammeter will not change the resistance appreciably and also the current in the circuit. for an ideal ammeter, the resistance must be equal to zero. Hence, th reading in ammeter is always lesser than the actual current in the circuit. Let `I_("ideal")` be current measured from ideal ammeter and `I_("actual")` be the actual current measured in the circuit by the ammeter . Then , the percentage error in measuring a current through an ammeter is
` (Delta I)/(I) xx 100% = (I_("ideal") - I_("actual"))/(I_("actual")) xx 100%`
(ii) Galvanometer to a voltmeter : A voltmeter is an instrument used to measure potential difference across any two points in the electrial circuits. it should not draw any current from the circuit otherwise the value of potential difference to be measured will change .
Voltmeter must have high resistance and when it is connected i parallel, it will not draw appreciable current so that it will indicate the true potential difference.
A galvanometer. is converted into a voltmeter by connecting high resistance `R_(h)` in series with galvanometer. the scale is now calibrated in volt and the range of voltmeter depends on the values of the resistance connected in series i.e., the value of resistance is so adjusted that only current `I_(g)` produces full scale deflection in the galvanometer.
Let `R_(g)` be the resistance of galvanometer and `I_(g)` be the current with which the galvanometer produces full scale deflection . since the galvanometer is connected in series with high resistance , the current in the electrical circuit is same as the current passing through the galvanometer.
I `= I_(g)`
I = `I_(g) rArr I_(g) = ("Potential difference ")/("total resistance ") `
Since the galvanometer and high resistance are connected in series, the total resistance or effective resistance gives the resistance of voltmeter. The voltmeter resistance is `R_(v) = R_(g) + R_(h)`
Therefore, `I_(g) = (V)/(R_(g) + R_(v)) rArr R_(h) = (V)/(I_(g)) - R_(g)`
Note that `I_(g) prop` V
the deflection in the galvanometer is proportional to current `I_(g)`. But corrent `I_(g)` is proportional to the potential difference. Hence the deflection in the galvanometer is proportional to potential difference. Since the resistance of voltmeter is very large, a voltmeter connected in an electrical circuit will draw least current in the circuit . An ideal voltmeter is one which has infinite resistance.
