
A $60\mu F$ capacitor is connected to a 110 V, 60 Hz a.c. supply. Determine the r.m.s value of current in the circuit.
Answer
560.1k+ views
Hint: RMS value mentioned in the question refers to the root mean square values (effective values). The given voltage can be considered as its rms value. Then using ohm’s law, we can calculate the rms value of current. As no resistor is connected, the resistance to the flow of current will be provided by the capacitor only.
Complete Step by step answer: Given:
Capacitance (C) of the capacitor = $60\mu F = 6 \times {10^6}F\left( {\because 1F = {{10}^6}\mu F} \right)$
Root mean square value of voltage = 110 V
Frequency $\left( \nu \right)$ = 60 Hz.
Now, according to ohm’s law:
V = IR but here we are considering root mean square values of both voltage and current, so:
$
{V_{rms}} = {I_{rms}}R \\
\Rightarrow {I_{rms}} = \dfrac{{{V_{rms}}}}{R}......(1) \\
$
Because we need to calculate the r.m.s value of current.
In the circuit, there is no resistor connected. So the net resistance only be provided by the capacitor known as capacitive reactance which is given as:
$
{X_C} = \dfrac{1}{{\omega C}} \\
\Rightarrow {X_C} = \dfrac{1}{{2\pi \nu C}}\left( {\because \omega = 2\pi \nu } \right) \\
$
Substituting the given values of frequency and capacitance, we get:
$
{X_C} = \dfrac{1}{{2\pi \nu C}} \\
\Rightarrow {X_C} = \dfrac{1}{{2 \times 3.14 \times 60 \times 60 \times {{10}^{ - 6}}}} \\
\Rightarrow {X_C} = 44.2\Omega \\
$
As this is the resistance provided by to the flow of current, its SI unit will be ohms $\left( \Omega \right)$
Equation (1) becomes:
${I_{rms}} = \dfrac{{{V_{rms}}}}{{{X_C}}}$
Substituting all the known values, we get:
$
{I_{rms}} = \left( {\dfrac{{110}}{{44.2}}} \right)A \\
\Rightarrow {I_{rms}} = 2.49A \\
$
Therefore, the r.m.s value of current in the circuit is 2.49 Amperes.
Note: Root mean square value is known as effective or virtual value of alternating current (AC). the amount of heating effect produced is equivalent to that produced by direct current (DC) Important SI Units of different quantities are:
Capacitance – Faraday (F), Voltage – Volts (V), Resistance – Ohm’s $\left( \Omega \right)$ , Frequency – Hertz (Hz) and Current – Ampere (A)
From the units only, we can get an idea about the quantity which is talked about in the question.
Complete Step by step answer: Given:
Capacitance (C) of the capacitor = $60\mu F = 6 \times {10^6}F\left( {\because 1F = {{10}^6}\mu F} \right)$
Root mean square value of voltage = 110 V
Frequency $\left( \nu \right)$ = 60 Hz.
Now, according to ohm’s law:
V = IR but here we are considering root mean square values of both voltage and current, so:
$
{V_{rms}} = {I_{rms}}R \\
\Rightarrow {I_{rms}} = \dfrac{{{V_{rms}}}}{R}......(1) \\
$
Because we need to calculate the r.m.s value of current.
In the circuit, there is no resistor connected. So the net resistance only be provided by the capacitor known as capacitive reactance which is given as:
$
{X_C} = \dfrac{1}{{\omega C}} \\
\Rightarrow {X_C} = \dfrac{1}{{2\pi \nu C}}\left( {\because \omega = 2\pi \nu } \right) \\
$
Substituting the given values of frequency and capacitance, we get:
$
{X_C} = \dfrac{1}{{2\pi \nu C}} \\
\Rightarrow {X_C} = \dfrac{1}{{2 \times 3.14 \times 60 \times 60 \times {{10}^{ - 6}}}} \\
\Rightarrow {X_C} = 44.2\Omega \\
$
As this is the resistance provided by to the flow of current, its SI unit will be ohms $\left( \Omega \right)$
Equation (1) becomes:
${I_{rms}} = \dfrac{{{V_{rms}}}}{{{X_C}}}$
Substituting all the known values, we get:
$
{I_{rms}} = \left( {\dfrac{{110}}{{44.2}}} \right)A \\
\Rightarrow {I_{rms}} = 2.49A \\
$
Therefore, the r.m.s value of current in the circuit is 2.49 Amperes.
Note: Root mean square value is known as effective or virtual value of alternating current (AC). the amount of heating effect produced is equivalent to that produced by direct current (DC) Important SI Units of different quantities are:
Capacitance – Faraday (F), Voltage – Volts (V), Resistance – Ohm’s $\left( \Omega \right)$ , Frequency – Hertz (Hz) and Current – Ampere (A)
From the units only, we can get an idea about the quantity which is talked about in the question.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

