Chapter 21, Problem 80QAP

To determine

(a)

The voltage and current amplitudes for the light bulb connected to an ac generator.

Answer to Problem 80QAP

For the bulb connected to the ac generator, the voltage amplitude is found to be 13 V and the current amplitude is 1.53 A.

Explanation of Solution

Given:

Magnetic field applied

  $B=0.225\text{ T}$

Number of turns

  $N=33$

Length of a side of the square coil

  $l=15.0\text{ cm}=15.0\times {10}^{-2}\text{m}$

Frequency of the generator

  $f=745\text{ rpm}=\frac{745}{60}\text{rps}$

Resistance of the bulb

  $R=8.50\Omega$

Formula used:

The voltage amplitude is given by the expression,

  $V_0=NBA\left(2\pi f\right)$

Here, A is the area of the square coil, given by,

  $A=l^2$

Therefore,

  $V_0=NB{l}^2\left(2\pi f\right)$......(1)

The current amplitude is given by,

  $i_0=\frac{V_0}{R}$......(2)

Calculation:

Substitute the known values of variables in equation (1) and calculate the voltage amplitude.

  $\begin{array}{c}{V}_0=NB{l}^2\left(2\pi f\right)\\ =\left(33\right)\left(0.225\text{ T}\right){\left(15.0\times {10}^{-2}\text{m}\right)}^2\left(2\right)\left(3.14\right)\left(\frac{745}{60}\text{rps}\right)\\ =13.03\text{ V}\end{array}$

Calculate the current amplitude by substituting the value of R and the calculated value of $V_0$ in equation (2).

  $\begin{array}{c}{i}_0=\frac{V_0}{R}\\ =\frac{13.03\text{ V}}{8.50\Omega }\\ =1.533\text{ A}\end{array}$

Conclusion:

Thus, for the bulb connected to the ac generator, the voltage amplitude is found to be 13 V and the current amplitude is 1.53 A.

To determine

(b)

The average rate at which heat is generated in the bulb connected to the ac generator.

Answer to Problem 80QAP

The average rate at which heat is generated in the bulb connected to the ac generator is 9.99 W.

Explanation of Solution

Given:

The current amplitude

  $i_0=1.533\text{ A}$

Resistance of the bulb

  $R=8.50\Omega$

Formula used:

The average rate at which heat is generated in the bulb is equal to the power dissipated by the bulb. This is given by,

  $P_{\text{av}}=i_{\text{rms}}^{2}R$

Here, $i_{\text{rms}}$ is the rms current flowing in the bulb and it is related to the current amplitude as,

  $i_{\text{rms}}=\frac{i_0}{\sqrt{2}}$

Therefore,

  $P_{\text{av}}=\frac{i_0^2}{2}R$......(3)

Calculation:

Substitute the values of the variables in equation (3) and calculate the rate at which heat is generated in the bulb.

  $\begin{array}{c}{P}_{\text{av}}=\frac{i_0^2}{2}R\\ =\frac{{\left(1.533\text{ A}\right)}^2}{2}\left(8.50\Omega \right)\\ =9.988\text{ W}\end{array}$

Conclusion:

Thus, the average rate at which heat is generated in the bulb connected to the ac generator is 9.99 W.

To determine

(c)

The energy consumed by the bulb every hour.

Answer to Problem 80QAP

The energy consumed by the bulb every hour is $3.6\times {10}^4\text{J}$.

Explanation of Solution

Given:

Average power

  $P_{\text{av}}=9.988\text{ W}$

Time interval

  $\Delta t=1\text{ h}$

Formula used:

The energy consumed by the bulb every hour is given by,

  $E=P_{\text{av}}\Delta t$

Calculation:

Substitute the values of variables in the formula and calculate the energy consumed in one hour.

  $\begin{array}{c}E=P_{\text{av}}\Delta t\\ =\left(9.988\text{ W}\right)\left(1\text{h}\times \frac{3600\text{ s}}{1\text{ h}}\right)\\ =3.595\times {10}^4\text{J}\\ =3.6\times {10}^4\text{J}\end{array}$

Conclusion:

Thus, the energy consumed by the bulb every hour is $3.6\times {10}^4\text{J}$.