The actual value of resistance R,shown in the | vedclass

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

Question

(B) Let $R = 30\,\Omega$ be the actual resistance and $R_v$ be the internal resistance of the voltmeter.In the given circuit,the voltmeter is connecte

Vedclass · Accepted Answer

$570$

Vedclass · Answer

(B) Let $R = 30\,\Omega$ be the actual resistance and $R_v$ be the internal resistance of the voltmeter.In the given circuit,the voltmeter is connected in parallel with the resistor $R$.The equivalent resistance $R_{eq}$ of the parallel combination is given by:$R_{eq} = \frac{R \cdot R_v}{R + R_v}$The measured value of resistance is $R_{measured} = R_{eq} = \frac{30 R_v}{30 + R_v}$.According to the problem,the measured value is $5\%$ less than the actual value:$R_{measured} = R 	imes (1 - 0.05) = 30 	imes 0.95 = 28.5\,\Omega$.Equating the two expressions for $R_{measured}$:$\frac{30 R_v}{30 + R_v} = 28.5$$30 R_v = 28.5(30 + R_v)$$30 R_v = 855 + 28.5 R_v$$1.5 R_v = 855$$R_v = \frac{855}{1.5} = 570\,\Omega$.Thus,the internal resistance of the voltmeter is $570\,\Omega$.

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.

The figure shows a network of resistances in which the point $A$ is earthed. The current through the $3\,\Omega$ resistor is

The total current supplied to the following circuit by the battery is (in $ A$)

In how much time will one litre of ${H_2}$ be collected by a $5\, A$ current? (Given: $Z = 1 \times {10^{ - 8}}\,kg/C$ and density of ${H_2} = 0.09\,kg/{m^3}$)

If the mass deposited in $30 \ s$ is $m$,what will be the $E.C.E$ of the copper voltameter?

Vedclass Test Series

Exam Paper Generator

Online Exam Module