If the resistance of the voltmeter is 10000 , | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Based on the circuit diagram,the voltmeter is connected in parallel to the series combination of the resistor $R$ and the ammeter.The reading of t

Vedclass · Accepted Answer

$118 \,\Omega$

Vedclass · Answer

(A) Based on the circuit diagram,the voltmeter is connected in parallel to the series combination of the resistor $R$ and the ammeter.The reading of the voltmeter represents the potential difference across the points $A$ and $B$,which is equal to the potential drop across the series combination of $R$ and the ammeter.Let $V = 12 \,V$ be the voltmeter reading and $I = 0.1 \,A$ be the ammeter reading.The total resistance between $A$ and $B$ is $R_{eq} = R + R_{ammeter} = R + 2 \,\Omega$.Using Ohm's law,$V = I 	imes R_{eq}$$12 = 0.1 	imes (R + 2)$$120 = R + 2$$R = 120 - 2 = 118 \,\Omega$.Thus,the value of $R$ is $118 \,\Omega$.

If the resistance of the voltmeter is $10000 \,\Omega$ and the resistance of the ammeter is $2 \,\Omega$,find the value of $R$ when the voltmeter reads $12 \,V$ and the ammeter reads $0.1 \,A$.

Similar Questions

In the circuit shown,the cell has $emf = 10\,V$ and internal resistance $= 1\, \Omega$. Find the current through the $4\, \Omega$ resistor.

Twelve wires each having resistance $2 \Omega$ are joined to form a cube. $A$ battery of $6 \text{ V}$ emf is connected across points $a$ and $c$. The voltage difference between $e$ and $f$ is . . . . . . $\text{V}$.

In the following circuit diagram,$E = 4\,V, r = 1\,\Omega$ and $R = 45\, \Omega$,then the reading of the ammeter $A$ will be

The current $i$ in the circuit shown in the figure is

In the figure shown,the capacity of the capacitor $C$ is $2\,\mu F$. The current in the $2\,\Omega$ resistor is ............... $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module