A resistor of 500Ω and one of 2000Ω are placed in series with a 60V supply. What will be the reading on a voltmeter of internal resistance 2000Ω when placed across (i) 500Ω resistor and (ii) 2000Ω resistor?

Respuesta :

Answer: 10 V, 40 V

Explanation:

Given

Resistors are [tex]500\Omega[/tex] and [tex]2000\Omega[/tex]

Power supply [tex]V=60\ V[/tex]

Voltmeter internal resistance [tex]r=2000\ \Omega[/tex]

When it connected around [tex]500\Omega[/tex] resistance, effective resistance becomes

[tex]\Rightarrow \dfrac{1}{R}=\dfrac{1}{500}+\dfrac{1}{2000}\\\\\Rightarrow R=\dfrac{2000}{5}\\\\\Rightarrow R=400\ \Omega[/tex]

Total resistance in series [tex]R_{net}=2400\ \Omega[/tex]

Current in circuit

[tex]I=\dfrac{60}{2400}=\dfrac{1}{40}\ A[/tex]

Voltage around [tex]500\Omega[/tex] resistance is

[tex]V=\dfrac{1}{40}\times 400\\\\V=10\ V[/tex]

(B) when voltmeter is connected around [tex]2000\Omega[/tex]  resistor

Net resistance around it

[tex]\Rightarrow \dfrac{1}{R}=\dfrac{1}{2000}+\dfrac{1}{2000}\\\\\Rightarrow R=1000\ \Omega[/tex]

Total resistance [tex]R_{net}=1500\ \Omega[/tex]

Current in circuit

[tex]I=\dfrac{60}{1500}=\dfrac{1}{25}\ A[/tex]

Voltage around [tex]2000\Omega[/tex] resistor

[tex]V=\dfrac{1}{25}\times 1000\\\\V=40\ V[/tex]