
Voltmeters \[{{V}_{1}}\] and \[{{V}_{2}}\] are connected in series across a D.C. line. \[{{V}_{1}}\] reads 80 V and has a per volt resistance of\[200\,\Omega \] , \[{{V}_{2}}\] has a total resistance of \[32\,\,k\Omega \] . The line voltage is:
A. 120 V
B. 160 V
C. 220 V
D. 240 V
Answer
533.1k+ views
Hint: Firstly we will find the value of the resistance of the voltmeter 1. Using this value of resistance and the voltage value of voltmeter 1, we will find the value of the current. Secondly, using this value of current, we will compute the value of the voltage reading of the voltmeter 2. Finally, using the voltage values of voltmeters 1 and 2, we will compute the value of the line voltage.
Formula used:
\[V=IR\]
Complete step-by-step answer:
From the given information, we have the data as follows.
The potential across the voltmeter \[{{V}_{1}}\], \[{{V}_{1}}=80\,V\]
The per volt resistance of the voltmeter \[{{V}_{1}}\], \[R=200\,\Omega \]
The resistance of the voltmeter \[{{V}_{2}}\], \[{{R}_{2}}=32\,\,k\Omega \]
Now, we will compute the total resistance of the voltmeter \[{{V}_{1}}\].
Consider the formula
\[{{R}_{1}}={{V}_{1}}\times R\]
Substitute the values in the above formula
\[\begin{align}
& {{R}_{1}}=80\times 200 \\
& \Rightarrow {{R}_{1}}=16\,k\Omega \\
\end{align}\]
Now, we will compute the current in the circuit.
Consider the formula
\[I=\dfrac{{{V}_{1}}}{{{R}_{1}}}\]
Substitute the values in the above formula
\[\begin{align}
& I=\dfrac{80}{16000} \\
& \Rightarrow I=5\,mA \\
\end{align}\]
Using this value of current, now we will compute the potential across the voltmeter \[{{V}_{2}}\].
Consider the formula
\[{{V}_{2}}=I{{R}_{2}}\]
Substitute the values in the above formula
\[\begin{align}
& {{V}_{2}}=5\times {{10}^{-3}}\times 32\times {{10}^{3}} \\
& \Rightarrow {{V}_{2}}=160\,V \\
\end{align}\]
The line voltage is equal to the sum of voltmeters present in a circuit.
Consider the formula for computing the value of the line voltage.
\[V={{V}_{1}}+{{V}_{2}}\]
Substitute the values in the above formula
\[\begin{align}
& V=80+160 \\
& \Rightarrow V=240\,V \\
\end{align}\]
So, the correct answer is “Option D”.
Note: The resistance per volt and the total resistance of the voltmeter are different values. The potential value is different from the line voltage, as the potential is across the single voltmeters, whereas, the line voltage is an algebraic sum of the potentials.
Formula used:
\[V=IR\]
Complete step-by-step answer:
From the given information, we have the data as follows.
The potential across the voltmeter \[{{V}_{1}}\], \[{{V}_{1}}=80\,V\]
The per volt resistance of the voltmeter \[{{V}_{1}}\], \[R=200\,\Omega \]
The resistance of the voltmeter \[{{V}_{2}}\], \[{{R}_{2}}=32\,\,k\Omega \]
Now, we will compute the total resistance of the voltmeter \[{{V}_{1}}\].
Consider the formula
\[{{R}_{1}}={{V}_{1}}\times R\]
Substitute the values in the above formula
\[\begin{align}
& {{R}_{1}}=80\times 200 \\
& \Rightarrow {{R}_{1}}=16\,k\Omega \\
\end{align}\]
Now, we will compute the current in the circuit.
Consider the formula
\[I=\dfrac{{{V}_{1}}}{{{R}_{1}}}\]
Substitute the values in the above formula
\[\begin{align}
& I=\dfrac{80}{16000} \\
& \Rightarrow I=5\,mA \\
\end{align}\]
Using this value of current, now we will compute the potential across the voltmeter \[{{V}_{2}}\].
Consider the formula
\[{{V}_{2}}=I{{R}_{2}}\]
Substitute the values in the above formula
\[\begin{align}
& {{V}_{2}}=5\times {{10}^{-3}}\times 32\times {{10}^{3}} \\
& \Rightarrow {{V}_{2}}=160\,V \\
\end{align}\]
The line voltage is equal to the sum of voltmeters present in a circuit.
Consider the formula for computing the value of the line voltage.
\[V={{V}_{1}}+{{V}_{2}}\]
Substitute the values in the above formula
\[\begin{align}
& V=80+160 \\
& \Rightarrow V=240\,V \\
\end{align}\]
So, the correct answer is “Option D”.
Note: The resistance per volt and the total resistance of the voltmeter are different values. The potential value is different from the line voltage, as the potential is across the single voltmeters, whereas, the line voltage is an algebraic sum of the potentials.
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