Two identical moving coil galvanometers have 10 resistance and full scale deflection at current. One of them is converted into a voltmeter of full scale reading and the other into an ammeter of full scale current using appropriate resistors. These are

Question

Two identical moving coil galvanometers have 10 resistance and full scale deflection at $2 μ A$ current. One of them is converted into a voltmeter of $100 mV$ full scale reading and the other into an ammeter of $1 mA$ full scale current using appropriate resistors. These are then used to measure the voltage and current in the Ohm's law experiment with $R = 1000 Ω$ resistor by using an ideal cell. Which of the following statement(s) is(are) correct? [2019]

Smart Achievers · Accepted Answer

(2, 4)

$Here, G = 10 Ω, I_{g} = 2 \times 10^{- 6} A, V = 100 mV = 0.1 V, I = 10^{- 3} A$

Using, $V = I_{g} (G + R)$ [R = resistance connected in series with the galvanometer]

$\Rightarrow 0.1 = 2 \times 10^{- 6} R_{v}$

$∴$ $R_{v} = 5 \times 10^{4} Ω$ (Resistance of voltmeter)

Also, $I_{g} G = (I - I_{g}) S$

$2 \times 10^{- 6} \times 10 = (10^{- 3} - 2 \times 10^{- 6}) S$

$∴$ $S = 2 \times 10^{- 2} Ω$ (Resistance of ammeter)

$R_{A} = \frac{G S}{G + S} = \frac{10 \times 0.02}{10 + 0.02} \approx 0.02 Ω$

$I = \frac{E}{\frac{50, 000 \times 1000}{51000} + 0.02} = \frac{E}{980.41}$

$V_{a b} = \frac{E}{980.41} \times \frac{50, 000 \times 1000}{51000} = \frac{E}{980.41} \times 980.39$

$∴$ $R_{measured} = \frac{V_{a b}}{I} = \frac{\frac{E}{980.41} \times 980.39}{\frac{E}{980.41}} = 980.39 Ω$

If the voltmeter shows full-scale deflection, then

$0.1 = \frac{E}{980} \times (\frac{1000}{51000}) \times 5 \times 10^{4}$

$∴$ $E = 999.6 mV$

Since $i_{A} = 10^{- 3} A$

$∴$ $Maximum reading of R = \frac{999.6 \times 10^{- 3}}{1 \times 10^{- 3}} = 999.6 Ω$

Solution

1 The resistance of the voltmeter will be $100 k$

2 The measured value of $R$ will be $978 < R < 982$

3 If the ideal cell is replaced by a cell having internal resistance of 5, then the measured value of $R$ will be more than 1000

4 The resistance of the ammeter will be 0.02 (round off to $2^{nd}$ decimal place).

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Solution

1 The resistance of the voltmeter will be 100k

2 The measured value of R will be 978<R<982

3 If the ideal cell is replaced by a cell having internal resistance of 5, then the measured value of R will be more than 1000

4 The resistance of the ammeter will be 0.02 (round off to 2nd decimal place).

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1 The resistance of the voltmeter will be $100 k$

2 The measured value of $R$ will be $978 < R < 982$

3 If the ideal cell is replaced by a cell having internal resistance of 5, then the measured value of $R$ will be more than 1000

4 The resistance of the ammeter will be 0.02 (round off to $2^{nd}$ decimal place).