Moving Coil Galvanometer — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Moving Coil Galvanometer MCQs with step-by-step solutions (4 questions). Part of Moving Charges and Magnetism. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Moving Coil Galvanometer · medium · numerical
In an ammeter $0.2\%$ of main current passes through the galvanometer. If resistance of galvanometer is $G$, the resistance of the ammeter will be
A. $\dfrac{1}{499}G$
B. $\dfrac{499}{500}G$
C. $\dfrac{1}{500}G$ ✓ Correct
D. $\dfrac{500}{499}G$
Solution: With $0.002I$ through the galvanometer $G$ and $0.998I$ through the shunt $r_s$: $0.002IG=0.998Ir_s \Rightarrow r_s\approx\dfrac{G}{499}$. The ammeter's equivalent resistance is $\dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{r_s}=\dfrac{1}{G}+\dfrac{499}{G}=\dfrac{500}{G}\Rightarrow R=\dfrac{G}{500}$.
Q2 — Moving Coil Galvanometer · medium · numerical
Current sensitivity of a moving coil galvanometer is $5\,\text{div/mA}$ and its voltage sensitivity (angular deflection per unit voltage applied) is $20\,\text{div/V}$. The resistance of the galvanometer is
A. $250\,\Omega$ ✓ Correct
B. $25\,\Omega$
C. $40\,\Omega$
D. $500\,\Omega$
Solution: Since $I_S=\dfrac{NAB}{k}$ and $V_S=\dfrac{NAB}{kR_G}=\dfrac{I_S}{R_G}$, $R_G=\dfrac{I_S}{V_S}=\dfrac{5\,\text{div/mA}}{20\,\text{div/V}}=0.25\,\text{V/mA}=250\,\Omega$.
Q3 — Moving Coil Galvanometer · medium · numerical
A millivoltmeter of $25\,\text{mV}$ range is to be converted into an ammeter of $25\,\text{A}$ range. The value (in ohm) of necessary shunt will be
A. $0.001$ ✓ Correct
B. $0.01$
C. $1$
D. $0.05$
Solution: Full-scale deflection current $i_g=\dfrac{25\,\text{mV}}{G}$. Required shunt $S=i_g\dfrac{G}{i}=\dfrac{25\,\text{mV}}{i}=\dfrac{25\times10^{-3}}{25}=0.001\,\Omega$.
Q4 — Moving Coil Galvanometer · medium · numerical
The resistance of an ammeter is $13\,\Omega$ and its scale is graduated for a current up to $100\,\text{A}$. After an additional shunt has been connected to this ammeter it becomes possible to measure currents up to $750\,\text{A}$ by this meter. The value of shunt resistance is
A. $20\,\Omega$
B. $2\,\Omega$ ✓ Correct
C. $0.2\,\Omega$
D. $2\,\text{k}\Omega$
Solution: Since the potential difference across the ammeter and shunt is equal, $S=\dfrac{i_aR}{i-i_a}$ with $i_a=100\,\text{A}$, $i=750\,\text{A}$, $R=13\,\Omega$: $S=\dfrac{100\times13}{750-100}=\dfrac{1300}{650}=2\,\Omega$.