Chapter 21, Problem 66P

(a)

To determine

The rms voltage across the lightbulb and the power dissipated by the lightbulb, if the lightbulb is on and hair dryer is off.

Answer to Problem 66P

The rms voltage across the lightbulb is $119.5\text{V}$ and the power dissipated by the lightbulb is $59.50\text{W}$.

Explanation of Solution

Apply Kirchhoff’s voltage rule in the circuit to get expression of rms current.

  $\begin{array}{c}0=\epsilon -I_{\text{rms}}r-I_{\text{rms }}{R}_1-I_{\text{rms}}r\\ I_{\text{rms}}=\frac{\epsilon }{2r+R_1}\end{array}$

Here, $r$ is the resistance of wiring in the walls, $R_1$ is the resistance of lightbulb and $I_{\text{rms}}$ is the rms current.

Write the expression for the rms voltage.

  $V_{\text{rms}}=I_{\text{rms}}{R}_1$

Here, $V_{\text{rms}}$ is the rms voltage, $I_{\text{rms}}$ is the rms voltage and $R_1$ is the resistance of lightbulb.

Substitute $\frac{\epsilon }{2r+R_1}$ for $I_{\text{rms}}$ in above equation to get $v_{\text{rms}}$.

  $V_{\text{rms}}=\frac{\epsilon R_1}{2r+R_1}$                                                                                                             (I)

Write the expression for the power dissipated by the lightbulb.

  $P=I_{\text{rms}}^2{R}_1$

Here, $P$ is the power dissipated by the lightbulb.

Substitute $\frac{\epsilon }{2r+R_1}$ for $I_{\text{rms}}$ in above equation to get $v_{\text{rms}}$.

  `$P=\frac{{\epsilon }^2}{{\left(2r+R_1\right)}^2}{R}_1$                                                                                                      (II)

Conclusion:

Substitute $120.0\text{V}$ for $\epsilon$, $240.0\Omega$ for $R_1$ and $0.50\Omega$ for $r$ in equation (I) to get $V_{\text{rms}}$.

  $\begin{array}{c}{V}_{\text{rms}}=\frac{\left(120.0\text{V}\right)\left(240.0\Omega \right)}{2\left(0.50\Omega \right)+\left(240.0\Omega \right)}\\ =119.5\text{V}\end{array}$

Substitute $120.0\text{V}$ for $\epsilon$, $240.0\Omega$ for $R_1$ and $0.50\Omega$ for $r$ in equation (II) to get $P$.

  $\begin{array}{c}P=\frac{{\left(120.0\text{V}\right)}^2\left(240.0\Omega \right)}{{\left[2\left(0.50\Omega \right)+\left(240.0\Omega \right)\right]}^2}\\ =59.50\text{W}\end{array}$

Therefore, the rms voltage across the lightbulb is $119.5\text{V}$ and the power dissipated by the lightbulb is $59.50\text{W}$.

(b)

To determine

The rms voltage across the lightbulb, the power dissipated by the lightbulb and rms voltage between point $A$ and ground, if both lightbulb and hair dryer is on.

Answer to Problem 66P

The rms voltage across the lightbulb is $111.0\text{V}$ , the power dissipated by the lightbulb is $51.34\text{W}$ and rms voltage between point $A$ and ground is $4.5\text{V}$.

Explanation of Solution

Write the expression for the rms current.

  $I_{\text{rms}}=\frac{\epsilon }{R_{\text{eq}}}$                                                                                                                  (III)

Here, $R_{\text{eq}}$ is the equivalent resistance of circuit.

Write the expression for the equivalent resistance of the circuit, if both light bulb and hairdryer are on.

  $R_{\text{eq}}=r+r+{\left(\frac{1}{R_1}+\frac{1}{r+r+R_2}\right)}^{-1}$                                                                                (IV)

Apply Kirchhoff’s voltage rule in the circuit to get expression of rms voltage across lightbulb.

  $\begin{array}{c}0=\epsilon -I_{\text{rms}}r-V_1-I_{\text{rms}}r\\ V_1=\epsilon -2I_{\text{rms}}r\end{array}$                                                                                             (V)

Here, $V_1$ is rms voltage across lightbulb.

Write the expression for the power dissipated by lightbulb.

  $P_1=\frac{V_{\text{1}}^2}{R_1}$                                                                                                                     (VI)

Here,$P_1$ is the power dissipated by lightbulb.

Write the expression for rms voltage between point $A$ to ground.

  $V_A=I_{\text{rms}}r$                                                                                                                 (VII)

Here, $V_A$ is the rms voltage between point $A$ to ground, $I_{\text{rms}}$ is the rms current through resistance $r$.

Conclusion:

Substitute $0.50\Omega$ for $r$ , $240.0\Omega$ for $R_1$ , $120.0\text{V}$ for $\epsilon$ and $12.0\Omega$ for $R_2$ in equation (IV) to get $R_{\text{eq}}$.

  $\begin{array}{c}{R}_{\text{eq}}=0.50\Omega +0.50\Omega +{\left(\frac{1}{240.0\Omega }+\frac{1}{0.50\Omega +0.50\Omega +12.0\Omega }\right)}^{-1}\\ =13.33\Omega \end{array}$

Substitute $120.0\text{V}$ for $\epsilon$ and $13.33\Omega$ for $R_{\text{eq}}$ in equation (III) to get $I_{\text{rms}}$.

  $\begin{array}{c}{I}_{\text{rms}}=\frac{120.0\text{V}}{13.33\Omega }\\ =9.00\text{A}\end{array}$

Substitute $120.0\text{V}$ for $\epsilon$ , $9.00\text{A}$ and $0.50\Omega$ for $r$ in equation (V) to get $V_1$.

  $\begin{array}{c}{V}_1=120.0\text{V}-2\left(9.00\text{A}\right)\left(0.50\Omega \right)\\ =111.0\text{V}\end{array}$

Substitute $111.0\text{V}$ for $V_1$ and $240.0\Omega$ for $R_1$ in equation (VI) to get $P_1$.

  $\begin{array}{c}{P}_1=\frac{{\left(111.0\text{V}\right)}^2}{240.0\Omega }\\ =51.34\text{W}\end{array}$

Substitute $9.00\text{A}$ for $I_{\text{rms}}$ and $0.50\Omega$ for $r$ in equation (VII) to get $V_A$.

  $\begin{array}{c}{V}_A=\left(9.00\text{A}\right)\left(0.50\Omega \right)\\ =4.5\text{V}\end{array}$

Therefore, the rms voltage across the lightbulb is $111.0\text{V}$ , the power dissipated by the lightbulb is $51.34\text{W}$ and rms voltage between point $A$ and ground is $4.5\text{V}$.

(c)

To determine

The explanation for why lights sometimes dim when an appliance is turned on.

Answer to Problem 66P

When appliances turned, it draws larger current for small time interval. This increase in current creates larger voltage drop across wires in the wall. This makes voltage drop across other appliances and other circuits in parallel to reduce from normal value. Thus makes lightbulb to dim when appliance is turned on.

Explanation of Solution

A lightbulb shows dimness when voltage drop across it is reduced. When an appliance is turned on, it draws large current for a brief time. The voltage drop across wires in the wall is proportional to the current flowing through them. Thus, when appliance is switched on, voltage drop across the wires in the wall increases and voltage drops across the appliances reduce.

Thus, while this large current flows, voltage drop across the wiring is larger than normal case, and the voltage drop across the appliance and across other circuit in parallel with it is smaller than usual.

(d)

To determine

The explanation for why neutral and the ground wires in a junction box are not at same potential even though they are both grounded.

Answer to Problem 66P

Neutral wires carry current whereas ground wires does not. Since a current flowing through neutral wire, there exists a potential difference between ends of neutral wire. Whereas grounded end always have a zero potential.

Explanation of Solution

In normal case, the no current flows through grounded wire. Potential at any grounded end is always zero. When current flows through the circuit, it flows through hot wire and into appliance, and in order to complete the circuit same current flows through the neutral wire.

This current produces voltage drop across resistance in the neutral wire. Since there is current flowing in the neutral wire, there is a potential difference $\Delta V=iR$ between the ends, where $\Delta V$ is the potential difference between the ends of neutral wire, $i$ is the current through the circuit and $R$ is the total resistance of neutral wire. Therefore, potential at the grounded ends are always at zero potential whereas the ends in the junction box are not.