A voltmeter has an internal resistance of 1000 Ω and gives full scale deflectioh when 2 V is applied across the terminals. Now a resistance of 4000 Ω is connected in series with it. Then it gives full scale deflection with

Question

A voltmeter has an internal resistance of 1000 Ω and gives full scale deflectioh when 2 V is applied across the terminals. Now a resistance of 4000 Ω is connected in series with it. Then it gives full scale deflection with (A) 8 V (B) 10 V (C) 6 V (D) 4 V

Tardigrade Admin · Accepted Answer

(V1/V2)=(R1/R2)=(2/V2)=(1000/1000+4000) Since Req = R 1+ R 2 V 2=10 V

Q. A voltmeter has an internal resistance of $1000 Ω$ and gives full scale deflectioh when 2 V is applied across the terminals. Now a resistance of $4000 Ω$ is connected in series with it. Then it gives full scale deflection with

8 V

10 V

6 V

4 V

$\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}} = \frac{2}{V _{2}} = \frac{1000}{1000 + 4000}$
Since $R e q = R_{1} + R_{2}$
$V_{2} = 10 V$

Q. A voltmeter has an internal resistance of 1000Ω and gives full scale deflectioh when 2 V is applied across the terminals. Now a resistance of 4000Ω is connected in series with it. Then it gives full scale deflection with

8 V

10 V

6 V

4 V

V2​V1​​=R2​R1​​=V2​2​=1000+40001000​ Since Req=R1​+R2​ V2​=10V

Q. A voltmeter has an internal resistance of $1000 Ω$ and gives full scale deflectioh when 2 V is applied across the terminals. Now a resistance of $4000 Ω$ is connected in series with it. Then it gives full scale deflection with

$\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}} = \frac{2}{V _{2}} = \frac{1000}{1000 + 4000}$
Since $R e q = R_{1} + R_{2}$
$V_{2} = 10 V$