Question

A \[100\Omega \] resistor is connected to a \[220V,50Hz\] supply.
(i) What is the rms value of current in the circuit?
(ii) What is the net power consumed over a full cycle?

Answer 1

Hint: Rms value of current is the effective value of the varying current, we use formula of rms current which is, \[{I_{rms}} = \dfrac{{{V_{rms}}}}{R}\]. After finding the rms value of current we will find the net power by using, \[P = \dfrac{{{V^2}}}{R}\].

Complete step by step answer:
Here \[{V_{rms}}\] is the rms value of the voltage and \[{I_{rms}}\] is the rms value of the current and R is the resistance.
Given-
\[{V_{rms}} = 220V\]
\[R = 100\Omega {\text{ }}\]
(i)Now, we use=\[{I_{rms}} = \dfrac{{{V_{rms}}}}{R}\]
Substitute values of \[R,{V_{rms}}\], we get-
\[{I_{rms}} = \dfrac{{220}}{{100}} = 2.2A\]
So, the root mean square value of the current is \[2.2A\]
(ii)Net power consumed over a full cycle,\[P = \dfrac{{{V^2}}}{R}\]
We know, v\[ = 220V\]
R\[ = 100\Omega \]
 Now put these two values in the equation,
We get- P\[ = {\dfrac{{\left( {220} \right)}}{{100}}^2}\]
\[P = \dfrac{{220 \times 220}}{{100}}\]
$\implies$ P\[ = 22 \times 22\]
$\therefore$ P\[ = 484W\]
So, the net power is\[ = 484W\], that is consumed over a full cycle.

Note:
Root mean square value of the current is the value of direct current that dissipates the same power in a resistor.
\[{I_{rms}} = \dfrac{{{V_{rms}}}}{R}\]
Root mean square value of voltage is equivalent voltage that represents DC voltage. Basically root mean square values are the square root of the mean or average values of the squared functions of the instantaneous values.
Electrical power can be time-varying either as a DC quantity or as an AC quantity and at any particular instant of time, is known as instantaneous power.