Physics, asked by bibekacharya524, 9 months ago

Two resistor of resistance 1000 ohm and 2000 ohm are joined in seris with a 100v supply . A voltmeter of internal resistance 4000 ohm is connected to measure the potential difference accross 1000 ohm resistor. Calculate the reading shown by the voltmeter.​

Answers

Answered by MisterIncredible
39

Given : -

Two resistors of resistance of 1000 ohm and 2000 ohm are joined in series with a 10V supply . A voltmeter of internal resistance of 4000 ohm is connected to measure the potential difference across 1000 ohm resistor .

Required to find : -

  • Calculate the reading shown by the voltmeter ?

Formula used : -

The formula to find the equivalent resistance when the resistors are connected in series .

\boxed{\tt{ R_{eq} = R_1 + R_2 + R_3 \dots \dots }}

The formula to find the equivalent resistance when the resistors are connected in parallel.

\boxed{\tt{ \dfrac{1}{ R_{eq} } = \dfrac{1}{ R_1 } + \dfrac{1}{ R_2 } + \dfrac{1}{ R_3} \dots \dots }}

Solution : -

Two resistors of resistance of 1000 ohm and 2000 ohm are joined in series with a 10V supply . A voltmeter of internal resistance of 4000 ohm is connected to measure the potential difference across 1000 ohm resistor .

we need to calculate the reading shown by the voltmeter ?

So,

First let's find the equivalent resistance of the whole circuit .

From the figure we can conclude that ;

The voltmeter of internal resistance 4000 ohm and the 1000 ohm resistor are connected in parallel . So, let's find their respective equivalent resistance .

Using the formula ;

\boxed{\tt{ \dfrac{1}{ R_{eq} } = \dfrac{1}{ R_1 } + \dfrac{1}{ R_2 } + \dfrac{1}{ R_3} \dots \dots }}

\tt{ \dfrac{1}{R_{eq}} = \dfrac{1}{ 1000 \Omega } + \dfrac{1}{4000 \Omega } }

\tt{ \dfrac{1}{ R_{eq}} = \dfrac{4 + 1 }{ 4000 \Omega } }

\tt{ \dfrac{1}{ R_{eq} } = \dfrac{ 5 }{ 4000 \Omega } }

\tt{ R_{eq} = \dfrac{ 4000 \Omega }{ 5 }}

\tt{ R_{eq} = 800 \Omega }

Hence,

Equivalent resistance between the voltmeter and 1000 ohm resistor is 800 ohm .

Now,

Let's find the equivalent resistance of the whole circuit .

Here,

The equivalent resistor which can replace the voltmeter and 1000 ohm resistor is 800 ohm resistor .

This 800 ohm resistor is in series with 2000 ohm resistor .

So,

This is the equivalent resistance of the whole circuit .

This implies ;

\boxed{\tt{ R_{eq} = R_1 + R_2 + R_3 \dots \dots }}

\tt{ R_{eq} = 800 \Omega + 2000 \Omega }

\tt{ R_{eq} = 2800 \Omega }

Hence,

Equivalent resistance of the whole circuit is 2800 ohm

Now,

Let's find the current obtained from the 100 volt supply source .

Using the Ohm's law ;

V = IR

100 = I x 2800

100 = 2800I

2800I = 100

I = 100/2800

I = 1/28 A

Hence,

Current in the whole circuit = 1/28 A

Now,

Let's find the reading in the voltmeter across 1000 ohm resistor.

Using the Ohm's law ,

V = IR

Here,

I = 1/28 A

R = 800 ohm

This implies ;

V = 1/28 + 800

V = 800/28

V = 28.5714. . . . .

V = 28.6 volts ( approximately )

Therefore,

Readings in the voltmeter which is connected across the 1000 ohm resistor = 28.6 volts ( approximately )

Attachments:
Answered by abdulrubfaheemi
2

Answer:

Given : -

Two resistors of resistance of 1000 ohm and 2000 ohm are joined in series with a 10V supply . A voltmeter of internal resistance of 4000 ohm is connected to measure the potential difference across 1000 ohm resistor .

Required to find : -

Calculate the reading shown by the voltmeter ?

Formula used : -

The formula to find the equivalent resistance when the resistors are connected in series .

\boxed{\tt{ R_{eq} = R_1 + R_2 + R_3 \dots \dots }}

R

eq

=R

1

+R

2

+R

3

……

The formula to find the equivalent resistance when the resistors are connected in parallel.

\boxed{\tt{ \dfrac{1}{ R_{eq} } = \dfrac{1}{ R_1 } + \dfrac{1}{ R_2 } + \dfrac{1}{ R_3} \dots \dots }}

R

eq

1

=

R

1

1

+

R

2

1

+

R

3

1

……

Solution : -

Two resistors of resistance of 1000 ohm and 2000 ohm are joined in series with a 10V supply . A voltmeter of internal resistance of 4000 ohm is connected to measure the potential difference across 1000 ohm resistor .

we need to calculate the reading shown by the voltmeter ?

So,

First let's find the equivalent resistance of the whole circuit .

From the figure we can conclude that ;

The voltmeter of internal resistance 4000 ohm and the 1000 ohm resistor are connected in parallel . So, let's find their respective equivalent resistance .

Using the formula ;

\boxed{\tt{ \dfrac{1}{ R_{eq} } = \dfrac{1}{ R_1 } + \dfrac{1}{ R_2 } + \dfrac{1}{ R_3} \dots \dots }}

R

eq

1

=

R

1

1

+

R

2

1

+

R

3

1

……

\tt{ \dfrac{1}{R_{eq}} = \dfrac{1}{ 1000 \text{\O}mega } + \dfrac{1}{4000 \text{\O}mega } }

R

eq

1

=

1000Ømega

1

+

4000Ømega

1

\tt{ \dfrac{1}{ R_{eq}} = \dfrac{4 + 1 }{ 4000 \text{\O}mega } }

R

eq

1

=

4000Ømega

4+1

\tt{ \dfrac{1}{ R_{eq} } = \dfrac{ 5 }{ 4000 \text{\O}mega } }

R

eq

1

=

4000Ømega

5

\tt{ R_{eq} = \dfrac{ 4000 \text{\O}mega }{ 5 }}R

eq

=

5

4000Ømega

\tt{ R_{eq} = 800 \text{\O}mega }R

eq

=800Ømega

Hence,

Equivalent resistance between the voltmeter and 1000 ohm resistor is 800 ohm .

Now,

Let's find the equivalent resistance of the whole circuit .

Here,

The equivalent resistor which can replace the voltmeter and 1000 ohm resistor is 800 ohm resistor .

This 800 ohm resistor is in series with 2000 ohm resistor .

So,

This is the equivalent resistance of the whole circuit .

This implies ;

\boxed{\tt{ R_{eq} = R_1 + R_2 + R_3 \dots \dots }}

R

eq

=R

1

+R

2

+R

3

……

\tt{ R_{eq} = 800 \text{\O}mega + 2000 \text{\O}mega }R

eq

=800Ømega+2000Ømega

\tt{ R_{eq} = 2800 \text{\O}mega }R

eq

=2800Ømega

Hence,

Equivalent resistance of the whole circuit is 2800 ohm

Now,

Let's find the current obtained from the 100 volt supply source .

Using the Ohm's law ;

V = IR

100 = I x 2800

100 = 2800I

2800I = 100

I = 100/2800

I = 1/28 A

Hence,

Current in the whole circuit = 1/28 A

Now,

Let's find the reading in the voltmeter across 1000 ohm resistor.

Using the Ohm's law ,

V = IR

Here,

I = 1/28 A

R = 800 ohm

This implies ;

V = 1/28 + 800

V = 800/28

V = 28.5714. . . . .

V = 28.6 volts ( approximately )

Therefore,

Readings in the voltmeter which is connected across the 1000 ohm resistor = 28.6 volts ( approximately )

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