Question

The voltage supplied by a wall socket varies with time,reversing its polarity with a constant frequency, as shown in thegraph. (Intro 1 figure)

Intro Fig.

Fig Part B

1.

What is the rms value $V_rms$ of the voltage plotted in the graph?

Express your answer in volts.

$V_rms$ = ? V

2.

When a lamp is connected to a wall plug, theresulting circuit can be represented by a simplified AC circuit, asshown in the figure. (Part B figure) Here the lamp has been replaced by a resistor with an equivalentresistance Part B figure) Here the lamp has been replaced bya resistor with an equivalent resistance = 120 $\Omega$ . What is the rms value $I_rms$ of the current flowing through the circuit?

Express your answer in amperes.

$I_rms$ = ? A

3.

What is the average power $P_avg$ dissipated in the resistor?

Express your answer in watts.

$P_avg$ = ? W

Answer 1

Concepts and reason

The concept used to solve this problem is root square mean value of current and voltage for AC circuits.

In the first part, use the relation between ${V_{rms}}$and peak value of voltage $\left( {{V_{peak}}} \right)$. Substitute the value of $\left( {{V_{peak}}} \right)$ and find the value of ${V_{rms}}$.

In the next part, use the relation of ${I_{rms}}$with ${V_{rms}}$and R and find the value of current.

Fundamentals

The root square mean value is defined as the square root of the mean value of the squared function of the instantaneous values. Mathematically, the expression is given as follows:

${V_{rms}} = \frac{{{V_{peak}}}}{{\sqrt 2 }}$

The expression of the ${I_{rms}}$ is given as follows:

${I_{rms}} = \frac{{{V_{rms}}}}{R}$

Here, R is the value of resistance.

(1)

The expression of ${V_{rms}}$is given as follows:

${V_{rms}} = \frac{{{V_{peak}}}}{{\sqrt 2 }}$

Substitute 170 V for ${V_{peak}}$ in the above equation.

$\begin{array}{c}\\{V_{rms}} = \frac{{170{\rm{ volts}}}}{{\sqrt 2 }}\\\\ = 120.21{\rm{ volts}}\\\end{array}$

(2)

Substitute 120.21 volts for V and $120\Omega$for R in the expression ${I_{rms}} = \frac{{{V_{rms}}}}{R}$.

$\begin{array}{c}\\{I_{rms}} = \frac{{120.21{\rm{ volts}}}}{{120\Omega }}\\\\ = 1.00{\rm{ A}}\\\end{array}$

Ans: Part 1

The value of the ${V_{rms}}$is 120.21 volts.

Part 2

The rms value of the flowing current is 1.0 A.