Class 12PhysicsMoving Charges and MagnetismThe Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion
EasyWhy should the resistance of an ammeter be as low as possible?
(N/A) An ammeter is connected in series with the component to measure the current flowing through it.
If the resistance of the ammeter is $R_{1}$ and the resistance of the circuit is $R$,then after connecting the ammeter,the total resistance of the circuit becomes $R + R_{1}$.
Consequently,the measured value of the current,$I = \frac{V}{R + R_{1}}$,will be less than the true value of the current that would flow in the absence of the ammeter $(I_{true} = \frac{V}{R})$.
To minimize this error and ensure that the ammeter measures the true value of the current,the resistance of the ammeter $(R_{1})$ should be as low as possible,ideally approaching zero.
If the resistance of the ammeter is $R_{1}$ and the resistance of the circuit is $R$,then after connecting the ammeter,the total resistance of the circuit becomes $R + R_{1}$.
Consequently,the measured value of the current,$I = \frac{V}{R + R_{1}}$,will be less than the true value of the current that would flow in the absence of the ammeter $(I_{true} = \frac{V}{R})$.
To minimize this error and ensure that the ammeter measures the true value of the current,the resistance of the ammeter $(R_{1})$ should be as low as possible,ideally approaching zero.
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Similar Questions
Two moving coil meters $M_1$ and $M_2$ have the following particulars:
$R_1 = 10\,\Omega, N_1 = 30, A_1 = 3.6 \times 10^{-3}\, m^2, B_1 = 0.25\, T$
$R_2 = 14\,\Omega, N_2 = 42, A_2 = 1.8 \times 10^{-3}\, m^2, B_2 = 0.50\, T$
(The spring constants are identical for the two meters). Determine the ratio of voltage sensitivity of $M_2$ and $M_1$.
$R_1 = 10\,\Omega, N_1 = 30, A_1 = 3.6 \times 10^{-3}\, m^2, B_1 = 0.25\, T$
$R_2 = 14\,\Omega, N_2 = 42, A_2 = 1.8 \times 10^{-3}\, m^2, B_2 = 0.50\, T$
(The spring constants are identical for the two meters). Determine the ratio of voltage sensitivity of $M_2$ and $M_1$.
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View Solution$A$ millivoltmeter of $25 \, mV$ range is to be converted into an ammeter of $25 \, A$ range. The value (in $\Omega$) of the necessary shunt will be:
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View SolutionThe resistance of an ammeter is $13\, \Omega$ and its scale is graduated for a current up to $100\, A$. After an additional shunt has been connected to this ammeter,it becomes possible to measure currents up to $750\, A$ by this meter. The value of the shunt resistance is:
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