The ammeter A reads 2, A and the voltmeter V | vedclass

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(C) In the given circuit,the ammeter $A$ is in series with the parallel combination of resistor $R$ and voltmeter $V$.Let $I_A = 2\, A$ be the current

Vedclass · Accepted Answer

More than $10\, \Omega$

Vedclass · Answer

(C) In the given circuit,the ammeter $A$ is in series with the parallel combination of resistor $R$ and voltmeter $V$.Let $I_A = 2\, A$ be the current read by the ammeter.Let $V_V = 20\, V$ be the voltage read by the voltmeter.The current $I_A$ splits into two paths: one through the resistor $R$ $(I_R)$ and one through the voltmeter $(I_V)$.So,$I_A = I_R + I_V = 2\, A$.The voltage across the parallel combination is $20\, V$,so $I_R = \frac{20}{R}$.Since the voltmeter has a finite resistance $R_V$,it draws some current $I_V = \frac{20}{R_V} > 0$.Therefore,$I_R = I_A - I_V = 2 - I_V Substituting $I_R = \frac{20}{R}$,we get $\frac{20}{R} This implies $R > \frac{20}{2}$,so $R > 10\, \Omega$.

The ammeter $A$ reads $2\, A$ and the voltmeter $V$ reads $20\, V$. The value of resistance $R$ is (Assuming finite resistances of ammeter and voltmeter).

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