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