Advertisements
Advertisements
प्रश्न
A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
Advertisements
उत्तर
Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection, Ig = 4 mA = 4 × 10−3 A
Range of the ammeter is 0, which needs to be converted to 6 A.
∴ Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
S = `("I"_"g""G")/("I" - "I"_"g")`
= `(4 xx 10^-3 xx 15)/(6 - 4 xx 10^-3)`
S = `(6 xx 10^-2)/(6 - 0.004)`
= `0.06/5.996`
≈ 0.01 Ω = 10 mΩ
Hence, a 10 mΩ shunt resistor is to be connected in parallel with the galvanometer.
APPEARS IN
संबंधित प्रश्न
A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R1 in series with the coil. If a resistance R2 is connected in series with it, then it can measures upto V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2
Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.
Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer?
A moving coil galvanometer has a resistance of 25Ω and gives a full scale deflection for a current of 10mA. How will you convert it into a voltmeter having range 0 - 100 V?
Why is it necessary to introduce a radial magnetic field inside the coil of a galvanometer?
Define current sensitivity of a galvanometer.
Why does a galvanometer when connected in series with a capacitor show a momentary deflection, when it is being charged or discharged?
How does this observation lead to modifying the Ampere's circuital law?
Hence write the generalised expression of Ampere's law.
Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter?
Draw a labelled diagram of a moving coil galvanometer and explain its working. What is the function of radial magnetic field inside the coil?
In the meter bridge experiment, balance point was observed at J with AJ = l.
(i) The values of R and X were doubled and then interchanged. What would be the new position of balance point?
(ii) If the galvanometer and battery are interchanged at the balance position, how will the alance point get affected?
A moving coil galvanometer can be converted into an ammeter by ______.
The coil of a moving coil galvanometer is wound over a metal frame in order to ______.
A galvanometer having a coil resistance of 60 Ω shows full-scale deflection when a current of 1.0 amp passes through it. It can be converted into an ammeter to read currents up to 5.0 amp by:
In an ammeter 0.5% of main current passes through galvanometer; If resistance of galvanometer is G, the resistance of ammeter will be.
When a galvanometer is shunted with a 4 Ω resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 Ω wire. The further reduction (find the ratio of decrease in current to the previous current) in the deflection will be (the main current remains the same)
A resistance of 3Ω is connected in parallel to a galvanometer of resistance 297Ω. Find the fraction of current passing through the galvanometer.
A voltmeter has a range of 0 - 20 V and a resistance of 500 Q. Explain how can be used to measure voltages from 0 - 200 volt?
A galvanometer of resistance 100 Ω gives a full-scale deflection for a potential difference of 200 mV.
- What must be the resistance connected to convert the galvanometer into an ammeter of the range 0-200 mA?
- Determine resistance of the ammeter.
