A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a resistance, it can be converted into a voltmeter of range 0 - 30 V. If connected to a resistance, it becomes an ammeter of range 0 - 1.5 A. The value of is

Question

A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a $4990 Ω$ resistance, it can be converted into a voltmeter of range 0 - 30 V. If connected to a $\frac{2 n}{249} Ω$ resistance, it becomes an ammeter of range 0 - 1.5 A. The value of $n$ is [2014]

Smart Achievers · Accepted Answer

(5)

From $(\frac{I - I_{g}}{I_{g}}) S = \frac{V}{I_{g}} - R$

$\frac{1.5 - 0.006}{0.006} \times \frac{2 n}{249} = \frac{30}{0.006} - 4990$

$∴$ $n = \frac{2490}{498} = 5$

Solution

Q.
A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a $4990 Ω$ resistance, it can be converted into a voltmeter of range 0 - 30 V. If connected to a $\frac{2 n}{249} Ω$ resistance, it becomes an ammeter of range 0 - 1.5 A. The value of $n$ is [2014]

Ans.
(5)

From $(\frac{I - I_{g}}{I_{g}}) S = \frac{V}{I_{g}} - R$

$\frac{1.5 - 0.006}{0.006} \times \frac{2 n}{249} = \frac{30}{0.006} - 4990$

$∴$ $n = \frac{2490}{498} = 5$

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Solution

Q. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω resistance, it can be converted into a voltmeter of range 0 - 30 V. If connected to a 2n249Ω resistance, it becomes an ammeter of range 0 - 1.5 A. The value of n is [2014]

Ans. (5) From (I-IgIg)S=VIg-R 1.5-0.0060.006×2n249=300.006-4990 ∴ n=2490498=5

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Q.
A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a $4990 Ω$ resistance, it can be converted into a voltmeter of range 0 - 30 V. If connected to a $\frac{2 n}{249} Ω$ resistance, it becomes an ammeter of range 0 - 1.5 A. The value of $n$ is [2014]

Ans.
(5)

From $(\frac{I - I_{g}}{I_{g}}) S = \frac{V}{I_{g}} - R$

$\frac{1.5 - 0.006}{0.006} \times \frac{2 n}{249} = \frac{30}{0.006} - 4990$

$∴$ $n = \frac{2490}{498} = 5$