
A voltmeter of least count 0.1V has a positive zero error of 5 divisions in a particular circuit. If it reads 25 divisions in a particular circuit, calculate the magnitude of potential difference. Also find out the resistance of the wire in ohm, used in the experiment to verify ohm’s law if the reading of the ammeter is 400 milliampere.
Answer
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Hint Obtain the correct reading first by finding the difference between the obtained reading on another circuit and the zero error division. Now using the correct reading, multiply with least count of the device to find magnitude of potential difference.
Complete Step By Step Solution
It is given that a voltmeter with a positive zero error of 5 divisions. Since it is a positive error, in order to obtain the correct value we need to subtract error division from obtained division value. Now, in our case, the obtained division value is given to be 25 divisions. Hence, the accurate reading will be ,
Accurate reading = Obtained reading – Error
\[ \Rightarrow Accurat{e_{}}Value = 25 - 5 = 20Divisions\]
Now, we know that all measuring devices consist of a least count. Least count of the measuring device is defined as the smallest and accurate value that can be measured by the device. In our voltmeter 0.1 V is the least count. Now, the magnitude of the reading can be found by the product of the number of accurate divisions and the least count.
\[V = Accurat{e_{}}Value \times L.C\]
\[ \Rightarrow V = 20 \times 0.1 = 2V\]
Now, Ohm’s law equation is given as\[V = I \times R\]. It is understood that this apparatus is used to verify ohm’s law. Thus, the current flowing in the circuit with a resistor of resistor R is given as\[400mA\].
The resistance of the material can be found by rearranging ohm’s law equation.
\[ \Rightarrow V = I \times R\]
\[ \Rightarrow R = \dfrac{V}{I}\]
\[ \Rightarrow R = \dfrac{2}{{400 \times {{10}^{ - 3}}}}\]
\[ \Rightarrow R = \dfrac{{2000}}{{400}}\]
\[ \Rightarrow R = 5\Omega \]
Therefore the resistance of the material is \[5\Omega \].
Note Ohm’s law is given by the statement “In a current carrying conductor , the current flowing through the material is directly proportional to the potential difference applied across the ends of the conducting material”.
Complete Step By Step Solution
It is given that a voltmeter with a positive zero error of 5 divisions. Since it is a positive error, in order to obtain the correct value we need to subtract error division from obtained division value. Now, in our case, the obtained division value is given to be 25 divisions. Hence, the accurate reading will be ,
Accurate reading = Obtained reading – Error
\[ \Rightarrow Accurat{e_{}}Value = 25 - 5 = 20Divisions\]
Now, we know that all measuring devices consist of a least count. Least count of the measuring device is defined as the smallest and accurate value that can be measured by the device. In our voltmeter 0.1 V is the least count. Now, the magnitude of the reading can be found by the product of the number of accurate divisions and the least count.
\[V = Accurat{e_{}}Value \times L.C\]
\[ \Rightarrow V = 20 \times 0.1 = 2V\]
Now, Ohm’s law equation is given as\[V = I \times R\]. It is understood that this apparatus is used to verify ohm’s law. Thus, the current flowing in the circuit with a resistor of resistor R is given as\[400mA\].
The resistance of the material can be found by rearranging ohm’s law equation.
\[ \Rightarrow V = I \times R\]
\[ \Rightarrow R = \dfrac{V}{I}\]
\[ \Rightarrow R = \dfrac{2}{{400 \times {{10}^{ - 3}}}}\]
\[ \Rightarrow R = \dfrac{{2000}}{{400}}\]
\[ \Rightarrow R = 5\Omega \]
Therefore the resistance of the material is \[5\Omega \].
Note Ohm’s law is given by the statement “In a current carrying conductor , the current flowing through the material is directly proportional to the potential difference applied across the ends of the conducting material”.
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