Question

A pure resistor & pure inductor is connected as shown in figure, now alternating voltage is connected across setup and reading of voltmeter is $V_1$ & reading of ammeter is $A_1$, now diamagnetic rod is placed in solenoid (inductor), now reading of voltmeter is $V_2$ & reading of ammeter is $A_2$, then

• A. $V_2>V_1,A_1>A_2$
• B. $V_2<V_1,A_1>A_2$
• C. $V_2>V_1,A_1<A_2$
• D. $V_2<V_1,A_1<A_2$

Answer 1

The correct option is D $V_2<V_1,A_1<A_2$

$Z=\sqrt{{\left(R\right)}^2+{\left(X_L\right)}^2}$
Also,
$I=\frac{V}{Z}$
Due to diamagnetic rod, L is decreased & $X_L=\omega L$ also decreases and hence $Z$ decreases.
Since voltage supply is constant, $I$ increases. Therefore, $A_1<A_2$.
Hence, voltage drop in resistance R ($V_R$) increases and since, overall drop of voltage remain same as voltage supply is constant and present in inductor and resistance only , therefore voltage drop across inductor L $\left(V_L\right)$ decreases.Therefore, $V_2<V_1$.