Question

[IIT- JEE (Advanced) - 2014]

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 2n/249 Ω resistance, it becomes an ammeter of range 0 − 1.5 A. The value of n is ?


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Answer 1

Answer:-

$\pink{\bigstar}$ The value of n is $\large\leadsto\boxed{\tt\purple{5}}$

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• Given:-

• I = 1.5 A

• Ig = 0.006 A

• R = 4990 Ω

• V = 30 V

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• Figure:-

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\put(1,5.1){\line(1,0){2.5}}\qbezier(4.5,5.1)(4.5,5.1)(5.5,5.1)\qbezier(5.5,5.1)(5.5,5.1)(5.7,5.4)\qbezier(5.7,5.4)(5.7,5.4)(5.9,4.8)\qbezier(5.9,4.8)(5.9,4.8)(6.1,5.4)\qbezier(6.1,5.4)(6.1,5.4)(6.3,4.8)\qbezier(6.3,4.8)(6.3,4.8)(6.5,5.4)\qbezier(6.5,5.4)(6.5,5.4)(6.7,4.8)\qbezier(6.7,4.8)(6.7,4.8)(6.8,5.1)\qbezier(6.8,5.1)(6.8,5.1)(10,5.1)\put(4,5.1){\circle{1}}\put(3.8,5.0){\bf G}\put(5.5,4.3){\bf r = 10 \: \Omega }\put(5.5,1.3){\bf \dfrac{2n}{249} \: \Omega }\put(10,5.1){\line(0,-1){3}}\put(1,5.1){\line(0,-1){3}}\qbezier(1,2.1)(1,2.1)(4.5,2.1)\qbezier(4.5,2.1)(4.5,2.1)(5.5,2.1)\qbezier(5.5,2.1)(5.5,2.1)(5.7,2.4)\qbezier(5.7,2.4)(5.7,2.4)(5.9,2.0)\qbezier(5.9,2.0)(5.9,2.0)(6.1,2.4)\qbezier(6.1,2.4)(6.1,2.4)(6.3,2.0)\qbezier(6.3,2.0)(6.3,2.0)(6.5,2.4)\qbezier(6.5,2.4)(6.5,2.4)(6.7,2.0)\qbezier(6.7,2.0)(6.7,2.0)(6.8,2.1)\qbezier(6.8,2.1)(6.8,2.1)(10,2.1)\put(1,5.1){\vector(1,0){1.5}}\put(2.1,4.4){\bf I_{g} }\end{picture}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Solution:-

✯ Let the initial resistance of galvanometer be 'r'.

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Now,

• When galvanometer is converted into Voltmeter:-

We know, a galvanometer can be converted into Voltmeter by connecting a large resistance across it in series.

Hence, we add the given Resistance R in series with Galvanometer.

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✯ Refer to the Figure:-

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Now,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{V = I_g(r+R)}}}$

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➪ $\sf 30 = 0.006 (r + 4990)$

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➪ $\sf r = \dfrac{30}{0.006} - 4990$

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➪ $\sf r = \dfrac{30 - 4990 \times 0.006}{0.006}$

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➪ $\sf r = \dfrac{30 - 29.94}{0.006}$

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➪ $\sf r = \dfrac{0.06}{0.006}$

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★ $\large{\underline{\underline{\bf\red{r = 10 \: \Omega}}}}$

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Therefore, the resistance of galvanometer 'r' is 10 Ω.

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• When Galvanometer is converted into Ammeter:-

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✯ Refer to the Figure:-

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$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{I_g \times r = (I - I_g) \times r}}}$

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➪ $\sf 0.006 \times 10 = (1.5 - 0.006) \times \dfrac{2n}{249}$

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➪ $\sf 0.06 = 1.494 \times \dfrac{2n}{249}$

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➪ $\sf 0.06 = \dfrac{2.988}{249} \times n$

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➪ $\sf n = \dfrac{0.06 \times 249}{2.988}$

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➪ $\sf n = \dfrac{14.94}{2.988}$

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★ $\large{\underline{\underline{\bf\red{n = 5}}}}$

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Therefore, the value of n is 5.

★ Note:- Reference of the exact figure in attachment.

Answer 2

Answer:

Given :-

• 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 2N/249 Ω resistance, it becomes an ammeter of range 0 - 15 A.

To Find :-

• What is the value of n.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{V = I_g(r + R)}}}$

$\longmapsto \sf\boxed{\bold{\pink{{I_g}r =\: (I - I_g)r}}}$

Solution :-

First, we have to find the resistance of a galvanometer :

Given :

• R = 4990 Ω
• V = 30 V
• Ig = 0.006 A
• I = 1.5 A

According to the question by using the formula we get,

$\implies \sf 30 =\: 0.006(r + 4990)$

$\implies \sf 30 =\: \dfrac{6}{1000}(r + 4990)$

$\implies \sf \dfrac{30 \times 1000}{6} =\: r + 4990$

$\implies \sf \dfrac{\cancel{30000}}{\cancel{6}} =\: r + 4990$

$\implies \sf \dfrac{5000}{1} =\: r + 4990$

$\implies \sf 5000 =\: r + 4990$

$\implies \sf 5000 - 4990 =\: r$

$\implies \sf 10 =\: r$

$\implies \sf\bold{\purple{r =\: 10\: \Omega}}$

Hence, the resistance of a galvanometer is 10 Ω .

Now, we have to find the value of n :

Given :

• Ig = 0.006 A
• r = 10 Ω
• I = 1.5 A
• r = 2n/249 Ω

According to the question by using the formula we get,

$\implies \sf 0.006 \times 10 =\: (1.5 - 0.006) \times \dfrac{2n}{249}$

$\implies \sf \dfrac{6}{1000} \times 10 =\: \bigg(\dfrac{15}{10} - \dfrac{6}{1000}\bigg) \times \dfrac{2n}{249}$

$\implies \sf \dfrac{60}{1000} =\: \bigg(\dfrac{1500 - 6}{1000}\bigg) \times \dfrac{2n}{249}$

$\implies \sf \dfrac{6\cancel{0}}{100\cancel{0}} =\: \dfrac{1494}{1000} \times \dfrac{2n}{249}$

$\implies \sf \dfrac{6}{100} =\: \dfrac{1494}{1000} \times \dfrac{2n}{249}$

$\implies \sf \dfrac{6}{100} =\: \dfrac{2988n}{249000}$

By doing cross multiplication we get,

$\implies \sf 298800n =\: 249000 \times 6$

$\implies \sf 298800n =\: 1494000$

$\implies \sf n =\: \dfrac{14940\cancel{00}}{2988\cancel{00}}$

$\implies \sf n =\: \dfrac{\cancel{14940}}{\cancel{2988}}$

$\implies \sf\bold{\red{n =\: 5}}$

$\therefore$ The value of n is 5 .