Class 12PhysicsMoving Charges and MagnetismThe Moving Coil Galvanometer (Sensitivity) and Ammeter and Voltmeter Conversion
MediumWhat is a shunt? Explain its function in a circuit,obtain the formula for the shunt,and explain its uses.
(N/A) Definition: $A$ shunt is a low-value resistor connected in parallel with a galvanometer to protect it and to increase its current-measuring range.
Function in Circuit:
$(i)$ Since the resistance of the shunt $(r_s)$ is much smaller than the galvanometer resistance $(R_G)$,most of the current passes through the shunt,protecting the galvanometer from damage due to high current.
$(ii)$ The equivalent resistance of the parallel combination is $R_{eq} = \frac{R_G r_s}{R_G + r_s}$. Since $R_G \gg r_s$,$R_{eq} \approx r_s$. This low resistance ensures that the original current in the circuit remains effectively unchanged.
Derivation of Formula:
Let $R_G$ be the resistance of the galvanometer and $I_G$ be the current flowing through it. Let $r_s$ be the shunt resistance and $(I - I_G)$ be the current flowing through it,where $I$ is the total current.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal:
$V_G = V_s$
$I_G R_G = (I - I_G) r_s$
$r_s = \frac{I_G R_G}{I - I_G}$
Uses:
$(i)$ It protects the galvanometer from damage by high currents.
$(ii)$ It is used to convert a galvanometer into an ammeter.
$(iii)$ By choosing an appropriate value of $r_s$,the range of the ammeter can be increased.
Function in Circuit:
$(i)$ Since the resistance of the shunt $(r_s)$ is much smaller than the galvanometer resistance $(R_G)$,most of the current passes through the shunt,protecting the galvanometer from damage due to high current.
$(ii)$ The equivalent resistance of the parallel combination is $R_{eq} = \frac{R_G r_s}{R_G + r_s}$. Since $R_G \gg r_s$,$R_{eq} \approx r_s$. This low resistance ensures that the original current in the circuit remains effectively unchanged.
Derivation of Formula:
Let $R_G$ be the resistance of the galvanometer and $I_G$ be the current flowing through it. Let $r_s$ be the shunt resistance and $(I - I_G)$ be the current flowing through it,where $I$ is the total current.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal:
$V_G = V_s$
$I_G R_G = (I - I_G) r_s$
$r_s = \frac{I_G R_G}{I - I_G}$
Uses:
$(i)$ It protects the galvanometer from damage by high currents.
$(ii)$ It is used to convert a galvanometer into an ammeter.
$(iii)$ By choosing an appropriate value of $r_s$,the range of the ammeter can be increased.
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Similar Questions
$A$ galvanometer having a coil resistance of $30\,\Omega$ shows full-scale deflection when a current of $2\,A$ passes through it. It can be converted into an ammeter to read currents up to $10\,A$ by:
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
View Solution$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series,showing a full-scale deflection of $30$ divisions. The additional series resistance required to reduce the deflection to $20$ divisions is: (in $\Omega$)
$A$ moving coil galvanometer is converted into an ammeter reading up to $0.03\,A$ by connecting a shunt of resistance $4r$ across it,and into an ammeter reading up to $0.06\,A$ when a shunt of resistance $r$ is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?
Medium
View Solution$A$ rectangular coil of effective area $0.05 \ m^2$ is suspended freely in a radial magnetic field of $0.01 \ Wb/m^2$. The torsional constant of the suspension fiber is $5 \times 10^{-9} \ Nm/\text{degree}$. If a current of $300 \ \mu A$ is passed through it,then the angle through which the coil rotates is (in $^{\circ}$)
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