Class 12PhysicsMoving Charges and MagnetismThe Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion
Medium$A$ galvanometer coil has a resistance of $12\; \Omega$ and the meter shows full-scale deflection for a current of $3\; mA$. How will you convert the meter into a voltmeter of range $0$ to $18\; V$?
(D) Resistance of the galvanometer coil,$G = 12\; \Omega$.
Current for which there is full-scale deflection,$I_{g} = 3\; mA = 3 \times 10^{-3}\; A$.
Range of the voltmeter,$V = 18\; V$.
To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
The formula for the series resistance is $R = \frac{V}{I_{g}} - G$.
Substituting the given values:
$R = \frac{18}{3 \times 10^{-3}} - 12$
$R = 6000 - 12 = 5988\; \Omega$.
Therefore,a resistor of $5988\; \Omega$ must be connected in series with the galvanometer.
Current for which there is full-scale deflection,$I_{g} = 3\; mA = 3 \times 10^{-3}\; A$.
Range of the voltmeter,$V = 18\; V$.
To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
The formula for the series resistance is $R = \frac{V}{I_{g}} - G$.
Substituting the given values:
$R = \frac{18}{3 \times 10^{-3}} - 12$
$R = 6000 - 12 = 5988\; \Omega$.
Therefore,a resistor of $5988\; \Omega$ must be connected in series with the galvanometer.
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The 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:
$A$ galvanometer has resistance '$G$' and range '$V_g$'. How much resistance is required to read voltage up to '$V$' volt?
If an ammeter is to be used in place of a galvanometer,then we must connect:
$A$ galvanometer has a resistance of $25 \, \Omega$ and a maximum of $0.01 \, A$ current can be passed through it. In order to change it into an ammeter of range $10 \, A$, the shunt resistance required is
Medium
View SolutionTwo moving coil meters,$M_{1}$ and $M_{2}$ have the following particulars:
$R_{1}=10 \,\Omega, \quad N_{1}=30$
$A_{1}=3.6 \times 10^{-3} \,m^{2}, \quad B_{1}=0.25 \,T$
$R_{2}=14 \,\Omega, \quad N_{2}=42$
$A_{2}=1.8 \times 10^{-3} \,m^{2}, \quad B_{2}=0.50 \,T$
(The spring constants are identical for the two meters). Determine the ratio of
$(a)$ current sensitivity and
$(b)$ voltage sensitivity of $M_{2}$ and $M_{1}$.
$R_{1}=10 \,\Omega, \quad N_{1}=30$
$A_{1}=3.6 \times 10^{-3} \,m^{2}, \quad B_{1}=0.25 \,T$
$R_{2}=14 \,\Omega, \quad N_{2}=42$
$A_{2}=1.8 \times 10^{-3} \,m^{2}, \quad B_{2}=0.50 \,T$
(The spring constants are identical for the two meters). Determine the ratio of
$(a)$ current sensitivity and
$(b)$ voltage sensitivity of $M_{2}$ and $M_{1}$.
Medium
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