An ammeter A of finite resistance and a resis | vedclass

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Question

(A) potentiometer is considered an ideal device that draws negligible current from the circuit,meaning its effective resistance is infinite. Thus,the

Vedclass · Accepted Answer

$I > I_0, V < V_0$

Vedclass · Answer

(A) potentiometer is considered an ideal device that draws negligible current from the circuit,meaning its effective resistance is infinite. Thus,the initial circuit behaves as if $R$ is connected directly to the cell $C$ through the ammeter $A$.When the potentiometer is replaced by a voltmeter of finite resistance $R_v$,the voltmeter acts as a resistor in parallel with $R$. The equivalent resistance of the parallel combination $(R || R_v)$ is less than $R$. Since the total resistance of the circuit decreases,the total current drawn from the ideal cell $C$ increases. Therefore,the ammeter reading $I$ is greater than the initial reading $I_0$ $(I > I_0)$.Because the ammeter has a finite resistance $R_A$,the voltage drop across the ammeter increases as the current increases. Since the cell $C$ is ideal (constant $EMF$ $E$),the voltage across the parallel combination $(R || R_v)$ is $V = E - I R_A$. Since $I > I_0$,it follows that $V < V_0$.

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(P)$ $M^1 L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(Q)$ $M^1 L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(R)$ $M^1 L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(S)$ None of these

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Match the following two columns:
Column-$I$ Column-$II$
$(A)$ Electrical resistance $(P)$ $M^1 L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential $(Q)$ $M^1 L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance $(R)$ $M^1 L^2 T^{-3} A^{-1}$
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