
A galvanometer having a coil resistance of $100\Omega$ gives a full scale deflection when a current of 1mA is passed through it. The value of the resistance which can convert this galvanometer into ammeter giving full scale deflection for a current of 10 A is
$\text{A.}\quad 0.01\Omega$
$\text{B.}\quad 2\Omega$
$\text{C.}\quad 0.1\Omega$
$\text{D.}\quad 3\Omega$
Answer
565.2k+ views
Hint: A galvanometer is an electrical instrument which detects the presence of current in the circuit. A galvanometer is a sensitive electromagnetic device which can measure low currents even of the order of a few microamperes. It consists of a metal piece on which wire is wound and a magnet field is applied, due to which it gets rotated on passing even the smallest current.
Formula used:
$V=IR$
Complete answer:
In this question, we are given that the galvanometer has a coil of resistance$100\Omega$. Also the galvanometer gives a full scale deflection when a current of 1mA is passed through it.
Hence the maximum potential difference that could be applied is:
$V = IR$
$V = 100\times 10^{-3} = 0.1 V$
Let the resistance of the shunt that is to be connected in parallel be ‘x’.
Thus the net resistance of the galvanometer will be:
$R_T = \dfrac{100x}{100+x}$ [parallel combination]
Hence corresponding to the same voltage and required value of current, resistance should be:
$R = \dfrac VI = \dfrac{0.1}{10} = 0.01 \Omega$
$\implies \dfrac{100x}{100+x} = 0.01$
$\implies 100x = 1 + 0.01x$
$\implies x \approx \dfrac{1}{100} = 0.01 \Omega$
Hence the shunt resistance should be $0.01\Omega$,
So, the correct answer is “Option A”.
Note:
A galvanometer is used to detect the presence of current in the circuit. It can be of many types. Some of them are tangent galvanometer, mirror galvanometer, astatic galvanometer, ballistic galvanometer and moving coil galvanometer. The most important and widely used galvanometer is the moving coil galvanometer. Students should always remember that the shunt should always be connected in parallel in order to convert it into ammeter and in series with the galvanometer in order to convert it into Voltmeter.
Formula used:
$V=IR$
Complete answer:
In this question, we are given that the galvanometer has a coil of resistance$100\Omega$. Also the galvanometer gives a full scale deflection when a current of 1mA is passed through it.
Hence the maximum potential difference that could be applied is:
$V = IR$
$V = 100\times 10^{-3} = 0.1 V$
Let the resistance of the shunt that is to be connected in parallel be ‘x’.
Thus the net resistance of the galvanometer will be:
$R_T = \dfrac{100x}{100+x}$ [parallel combination]
Hence corresponding to the same voltage and required value of current, resistance should be:
$R = \dfrac VI = \dfrac{0.1}{10} = 0.01 \Omega$
$\implies \dfrac{100x}{100+x} = 0.01$
$\implies 100x = 1 + 0.01x$
$\implies x \approx \dfrac{1}{100} = 0.01 \Omega$
Hence the shunt resistance should be $0.01\Omega$,
So, the correct answer is “Option A”.
Note:
A galvanometer is used to detect the presence of current in the circuit. It can be of many types. Some of them are tangent galvanometer, mirror galvanometer, astatic galvanometer, ballistic galvanometer and moving coil galvanometer. The most important and widely used galvanometer is the moving coil galvanometer. Students should always remember that the shunt should always be connected in parallel in order to convert it into ammeter and in series with the galvanometer in order to convert it into Voltmeter.
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