Chapter 19, Problem 17Q

Given the circuit shown in Fig. 19-38, use the words “increases,” “decreases,”or ”stays the same” to complete the following statements:

a. If R₇increases, the potential difference between A and E____ . Assume no resistancein and E.
b. If R₇-increases, the potential difference between A and E_____. Assume and Ehaveresistance.
c. If R₇ increases, the voltage drop across R_{4_____}
d. If R₂ decreases, the current through R_{1_______}
e. If R₂ decreases, the current through R_{6______}
f. If R₂ decreases, the current through R₃_____
g. If R₅ increases, the voltage drop across R_{2________}
h. If R₅ increases, the voltage drop across R_{4 _________}
I. If R_2, R_5, and R₇ increase, E (r=0)______

To determine

(a) To determine:

Whether the potential difference between point A and E increases, decreases or stays the same by increasing the value of resistance$R_7$ and assume there is no resistance between point A and E.

Answer to Problem 17Q

Solution:

The potential difference between point A and E increases by increasing the value of resistance$R_7$ if the value of resistance between point A and E is zero.

Explanation of Solution

Given info:

The circuit diagram of the given situation is shown below:

In the given circuit, the resistance $R_7$ is variable, means the magnitude of this resistance can be change. Since the resistance $R_7$ is connected in series with the circuit so, if the value of this resistance increases, the net equivalent resistance in the circuit increases.

From

Formula to calculate the Potential difference between points A and E from Ohm’s law is,

        $V=I{R}_{\text{eq}}$                                (I)

     •  $V$ is the potential difference between points $A$ and $E$.

     •  $I$ is the current flowing in the circuit.

     •  $R_{\text{eq}}$ is the net equivalent resistance in the circuit.

The net equivalent resistance from the circuit is,

      $R_{\text{eq}}=R_1+\left(\frac{1}{R_2}+\frac{1}{R_3}\right)+\left(\frac{1}{R_4}+\frac{1}{R_5}+\frac{1}{R_6}\right)+R_7$

Substitute $R_1+\left(\frac{1}{R_2}+\frac{1}{R_3}\right)+\left(\frac{1}{R_4}+\frac{1}{R_5}+\frac{1}{R_6}\right)+R_7$ for $R_{\text{eq}}$ in equation (I).

      $V=I\left[R_1+\left(\frac{1}{R_2}+\frac{1}{R_3}\right)+\left(\frac{1}{R_4}+\frac{1}{R_5}+\frac{1}{R_6}\right)+R_7\right]$

From the above equation, if the value of resistance $R_7$ increases the potential difference between the point A and E increases.

To determine

(b) To determine:

Whether the potential difference between point A and E increases, decreases or stays the same by increasing the value of resistance$R_7$ and assume the resistance between point A and E have some value.

Answer to Problem 17Q

Solution:

The potential difference between point A and E increases by increasing the value of resistance$R_7$ if the point A and E have some resistance.

Explanation of Solution

Suppose the resistance between the point A and E is $R$. This resistance connects in the circuit in series. Now, the net resistance in the circuit becomes $R_{\text{eq}}+R$.

Formula to calculate the Potential difference between points A and E from Ohm’s law is,

    $V=I\left(R_{\text{eq}}+R\right)$

Substitute $R_1+\left(\frac{1}{R_2}+\frac{1}{R_3}\right)+\left(\frac{1}{R_4}+\frac{1}{R_5}+\frac{1}{R_6}\right)+R_7$ for $R_{\text{eq}}$.

      $V=I\left[R_1+\left(\frac{1}{R_2}+\frac{1}{R_3}\right)+\left(\frac{1}{R_4}+\frac{1}{R_5}+\frac{1}{R_6}\right)+R_7+R\right]$

From the above equation, if the value of resistance $R_7$ is increases the potential difference between the point A and E increases.

To determine

(c) To determine:

Whether the voltage drop across resistance $R_4$ increases, decreases or stays the same by increasing the value of resistance$R_7$.

Answer to Problem 17Q

Solution:

The voltage drop across resistance $R_4$ decreases by increasing the value of resistance$R_7$.

Explanation of Solution

Formula to calculate the voltage drop across resistance $R_4$ is,

      $V_4=I{R}_4$

       • $V_4$ is the voltage drop across resistance $R_4$.

If the magnitude of the resistance $R_7$ is increases, the net resistance in the circuit increase that decreases the magnitude of total current flowing in the circuit. So, decrease in the value of total current the voltage drop across resistance $R_4$ decreases.

To determine

(d) To determine:

Whether the current through resistance $R_1$ increases, decreases or stays the same by decreasing the value of resistance$R_2$.

Answer to Problem 17Q

Solution:

The current through resistance $R_1$ increases by decreasing the value of resistance$R_2$.

Explanation of Solution

In the given circuit, the resistance $R_2$ and $R_3$ are connected in parallel between points B and C. So, if the resistance$R_2$ decreases the equivalent resistance between B and C decreases that further decreases the net equivalent resistance $R_{\text{eq}}$ in the circuit.

From the ohm’s law, if the net resistance in the circuit decreases for a given potential difference, the value of current flowing increases. So, the value of current through resistance $R_1$ increases.

To determine

(e) To determine:

Whether the current through resistance $R_1$ increases, decreases or stays the same by decreasing the value of resistance$R_2$.

Answer to Problem 17Q

Solution:

The current through resistance $R_6$ increases by decreasing the value of resistance$R_2$.

Explanation of Solution

In the given circuit, the resistance $R_2$ and $R_3$ are connected in parallel between points B and C. So, if the resistance$R_2$ decreases the equivalent resistance between B and C decreases that further decreases the net equivalent resistance $R_{\text{eq}}$ in the circuit.

From the ohm’s law, if the net resistance in the circuit decreases for a given potential difference, the value of current flowing increases. So, the value of current through resistance $R_6$

increases.

To determine

(f) To determine:

Whether the current through resistance $R_3$ increases, decreases or stays the same by decreasing the value of resistance$R_2$.

Answer to Problem 17Q

Solution:

The current through resistance $R_3$ decreases by decreasing the value of resistance$R_2$.

Explanation of Solution

Formula to calculate the potential difference across resistance $R_2$ is,

         $V_2=I_2{R}_2$

      • $V_2$ is the potential difference across resistance $R_2$.

      • $I_2$ is the current flow through resistance $R_2$.

Formula to calculate the potential difference across resistance $R_3$ is,

         $V_3=I_3{R}_3$

      • $V_3$ is the potential difference across resistance $R_3$.

      • $I_3$ is the current flow through resistance $R_3$.

The resistance $R_2$ and $R_3$ are connected in parallel between points B and C. The current flow will be different in both the resistance but the potential difference across remains same across both resistances.

        $V_2=V_3$

Substitute $I_2{R}_2$ for $V_2$ and $I_3{R}_3$ for $V_3$.

      $\begin{array}{c}{I}_2{R}_2=I_3{R}_3\\ I_3=\frac{R_2}{R_3}{I}_2\end{array}$

From the above expression if the value of resistance $R_2$ decreases, the value of current flowing through resistance $R_3$ decreases.

To determine

(g)To determine:

Whether the voltage drop across resistance $R_2$ increases, decreases or stays the same by increasing the value of resistance$R_5$.

Answer to Problem 17Q

Solution:

The voltage drop across resistance $R_2$ decreases by increasing the value of resistance$R_5$.

Explanation of Solution

In the given circuit, the resistance$R_4$, $R_5$ and $R_6$ are connected in parallel between points C and D. So, if the resistance$R_5$ increases the equivalent resistance between C and D increases that further increases the net equivalent resistance $R_{\text{eq}}$ in the circuit.

Formula to calculate the voltage drop across resistance $R_2$ is,

        $V_2=I{R}_2$

      • $V_2$ is the voltage drop across resistance $R_2$.

Hence, increases in the net equivalent resistance $R_{\text{eq}}$ the total value of current $I$ decreases so, the voltage drop across resistance $R_2$ decreases.

To determine

(h)To determine:

Whether the voltage drop across resistance $R_4$ increases, decreases or stays the same by increasing the value of resistance$R_5$.

Answer to Problem 17Q

Solution:

The voltage drop across resistance $R_4$ increases by increasing the value of resistance$R_5$.

Explanation of Solution

In the given circuit, the resistance$R_4$

, $R_5$ and $R_6$ are connected in parallel between points C and D. So, if the resistance$R_5$ increases the equivalent resistance between C and D increases that further increases the net equivalent resistance $R_{\text{eq}}$ in the circuit.

Formula to calculate the voltage drop across resistance $R_4$ is,

         $V_4=I{R}_4$

      • $V_4$ is the voltage drop across resistance $R_4$.

Formula to calculate the potential difference across resistance $R_5$ is,

         $V_5=I_5{R}_5$

      • $V_5$ is the potential difference across resistance $R_5$.

      • $I_5$ is the current flow through resistance $R_5$.

The resistance$R_4$, $R_5$ and $R_6$ are connected in parallel between points C and D. The current flow will be different in all the resistance but the potential difference across remains same across both resistances.

         $V_4=V_5$

Substitute $I_4{R}_4$ for $V_4$ and $I_5{R}_5$ for $V_5$.

      $\begin{array}{c}{I}_4{R}_5=I_5{R}_5\\ I_4=\frac{R_5}{R_4}{I}_5\end{array}$

From the above expression if the value of resistance $R_5$ increase, the value of current flows through resistance $R_4$ increase. So, the potential difference across the resistance $R_4$ increases.

To determine

(i) To determine:

Whether the emf $\xi$ of the battery increases, decreases or stays the same by increasing the value of resistances$R_5$, $R_2$ and $R_7$.

Answer to Problem 17Q

Solution:

The emf $\xi$ of the battery stays the same by increasing the value of resistances$R_5$, $R_2$ and $R_7$.

Explanation of Solution

In the given circuit, the resistance if the value of $R_2$, $R_5$ and $R_7$ are increases. The net equivalent resistance equivalent resistance $R_{\text{eq}}$ in the circuit increases. But the emf of the battery is depend upon the potential difference across battery and the internal resistance of the battery.

Formula to calculate the emf $\xi$ of the battery is,

        $\xi =V-Ir$

      • $\xi$ is the emf of the battery.

      • $r$ is the internal resistance.

Substitute $0$ for $r$.

      $\begin{array}{c}\xi =V-I\times 0\\ =V\end{array}$

The potential difference across the battery remains constant whatever the resistance of the circuit. So, the emf of the battery remains constant.