SATHEE: Current Electricity 4 Question 27

Current Electricity 4 Question 27

27. 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 Adv.)

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Solution:

$i_{g} (G + 4990) = V$

$\begin{aligned} \Rightarrow \frac{6}{1000} (G + 4990) = 30 \\ \Rightarrow G + 4990 = \frac{30000}{6} = 5000 \\ \Rightarrow G = 10 Ω \\ V_{a b} = V_{c d} \\ \Rightarrow i_{g} G = (1.5 - i_{g}) S \\ \Rightarrow \frac{6}{1000} \times 10 = 1.5 - \frac{6}{1000} S \\ \Rightarrow S = \frac{60}{1494} = \frac{2 n}{249} \\ \Rightarrow n = \frac{249 \times 30}{1494} = \frac{2490}{498} = 5 \end{aligned}$

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Current Electricity 4 Question 27

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

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27. 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