Question

Resistance of an ideal Ammeter is zero and that of an ideal voltmeter is infinite. A voltmeter and an ammeter are connected in the circuit as shown. Resistance of ammeter is say \[\dfrac{R}{10}\] and that of voltmeter is 10 R. Then:

A. Percentage error in the reading of Ammeter (compared to that measured, if both ammeter and voltmeter were ideal) is 1.0%.
B. Percentage error in the reading of voltmeter is 10.0%
C. Both (a) and (b) are correct
D. Both (a) and (b) are wrong

Answer 1

Hint: We have to find out the actual and observed value with the parameters given in the question. The ratio of the difference between the actual and observed value to the actual value is the error. From this, we can find out the percentage error.

The formula used: \[I=\dfrac{V}{R}\], where V is the voltage and R is the resistance. Here E is the voltage supplied to the circuit.

Complete step-by-step answer:
As we can see the circuit, in which the voltmeter is connected parallel to the resistance R and ammeter is connected in series with the resistor. From the circuit, we can find out the actual current flowing through the circuit.
\[I=\dfrac{V}{R}\], where V is the voltage and R is the resistance. Here E is the voltage supplied to the circuit.
\[I=\dfrac{E}{R}\]………………………(1)
Since the voltmeter is in parallel connection with the resistance R. The effective resistance at the resistor will be,
\[\dfrac{1}{{{R}_{v}}}=\dfrac{1}{R}+\dfrac{1}{10R}\], Where 10R is the resistance of voltmeter.
\[{{R}_{v}}=\dfrac{10R}{11}\]…………………………(2)
If we are using an ideal ammeter and ideal voltmeter, the current will be,
\[I=\dfrac{E}{{{R}_{a}}+{{R}_{v}}}\], where \[{{R}_{v}}\] is the resistance of voltmeter and \[{{R}_{a}}\] is the resistance of ammeter.
We can assign the obtained values into this equation.
\[I=\dfrac{E}{\dfrac{R}{10}+\dfrac{10R}{11}}\]
\[I=\dfrac{E}{\dfrac{111R}{110}}\]
\[I=\dfrac{110E}{111R}\]
\[I=0.99\dfrac{E}{R}\]………………………(3)
We are using an ammeter to measure the current. So now we can find out the error in the measurement of current.
The percentage error can be written as the ratio of the difference in actual and measured value to the actual value.
\[I=\left[ \dfrac{\dfrac{E}{R}-0.99\dfrac{E}{R}}{\dfrac{E}{R}} \right]\times 100\]
\[I=\left[ \dfrac{1-0.99}{1} \right]\times 100\]
\[I=0.01\times 100=1%\]
Next, we can find out the actual voltage through the R.
\[E=IR\]………………………(4)
The measured resistance is already shown in the equation (3). So, the measured voltage is,
\[V=I\left[ \dfrac{10R}{11} \right]\]
\[V=0.9IR\]
According to equation (4), \[E=IR\]
\[V=0.9E\]…………………….(5)
From this, we can calculate the percentage of error in the measurement of the voltmeter.
\[\left[ \dfrac{E-0.9E}{E} \right]\times 100=\left[ \dfrac{1-0.9}{1} \right]\times 100\]
\[0.1\times 100=10%\]
So, the correct option is C.

Note: Current measuring devices have to be in series with the circuit. Since the current measuring devices will add resistance to the circuit. So, it will reduce the measuring current than the actual current. That’s why we are preferring ideal ammeters with zero or negligible resistance.
Voltage measuring devices will connect in parallel to the circuit. It should have infinite resistance. Otherwise, a large amount of current flows through the voltage measuring device and will provide the wrong value. If we are using an ideal voltmeter with infinite or large resistance, we can measure the actual potential difference effectively.
Do not forget to multiply the ratio of error with 100. Then only it will become the percentage error.