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Question

(B) In the given circuit,the ammeter measures the total current $I_{total} = 4\, A$. The voltmeter is connected in parallel with the resistance $R$. L

Vedclass · Accepted Answer

more than $5\, \Omega$

Vedclass · Answer

(B) In the given circuit,the ammeter measures the total current $I_{total} = 4\, A$. The voltmeter is connected in parallel with the resistance $R$. Let the current flowing through the voltmeter be $i$. Then,the current flowing through the resistance $R$ is $(4 - i)$. The voltage across the resistance $R$ is given by the voltmeter reading,which is $20\, V$. Using Ohm's law for the resistance $R$: $V = I_R 	imes R$ $20 = (4 - i) 	imes R$ $R = \frac{20}{4 - i}$ Since the voltmeter is not ideal,it must draw some current $i > 0$. If $i > 0$,then $(4 - i) Therefore,$R = \frac{20}{4 - i} > \frac{20}{4} = 5\, \Omega$. Thus,the resistance $R$ is more than $5\, \Omega$.

$A$ student connects a voltmeter,ammeter,and resistance according to the circuit given. If the voltmeter reading is $20\, V$ and the ammeter reading is $4\, A$,then the resistance $R$ will be:

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