Question 594: Current Electricity

Question

A milliammeter of resistance $40 \Omega$ has a range $0-30 \mathrm{~mA}$. What will be the resistance used in series to convert it into voltmeter of range $0-15 \mathrm{~V}$ ?

Accepted Answer

To convert a milliammeter into a voltmeter with a range of $0-15 \, 	ext{V}$, the following information is given:
Maximum Voltage, $V = 15 \, 	ext{V}$
Current through the galvanometer, $I_g = 30 \, 	ext{mA} = 30 	imes 10^{-3} \, 	ext{A}$
Internal resistance of the galvanometer, $G = 40 \, \Omega$
We need to determine the resistance $R$ that should be added in series. The relationship for the voltmeter is given by:
$ V = I_g (R + G) $
Substituting the known values:
$ 15 = 30 	imes 10^{-3} (R + 40) $
Solving for $R$:
$ \begin{align*} 15 &= 30 	imes 10^{-3} (R + 40) \ 15 &= 0.03 (R + 40) \ \frac{15}{0.03} &= R + 40 \ 500 &= R + 40 \ R &= 500 - 40 = 460 \, \Omega \end{align*} $
Therefore, the resistance that should be used in series is $460 \, \Omega$.

Answer

$460 \Omega$

Answer

$920 \Omega$

Answer

$640 \Omega$

Answer

$560 \Omega$

Current Electricity

A milliammeter of resistance 40 \Omega has a range 0-30 \mathrm{~mA}. What will be the resistance used in series to convert it into voltmeter of range 0-15 \mathrm{~V} ?

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