Chapter 27, Problem 15P

Four resistors are connected to a battery as shown in Figure P27.15. (a) Determine the potential difference across each resistor in terms of ε. (b) Determine the current in each resistor in terms of I. (c) What If? If R₃ is increased, explain what happens to the current in each of the resistors. (d) In the limit that R₃ → ∞, what are the new values of the current in each resistor in terms of I, the original current in the battery?

Figure P27.15

To determine

The potential difference across each resistor in terms of $\epsilon$.

Answer to Problem 15P

The potential difference across $R_1$ resistor is $\frac{\epsilon }{3}$, potential difference across $R_2$ resistor is $\frac{2\epsilon }{9}$, potential difference across $R_3$ resistor is $\frac{4\epsilon }{9}$, potential difference across $R_4$ resistor is $\frac{2\epsilon }{3}$.

Explanation of Solution

The resistors $R_2$ and $R_3$ are connected in series combination as shown in figure 1.

Figure (1)

Formula to calculate the resistance across the circuit when resistors $R_2$ and $R_3$ are connected in series combination.

$R_S=R_2+R_3$ (1)

Here,

$R_S$ is the resistance across the circuit when resistors $R_2$ and $R_3$ are connected in series combination.

$R_2$ is the value of resistor 2.

$R_3$ is the value of resistor 3.

Substitute $2R$ for $R_2$, $4R$ for $R_3$ in equation (1) to find $R_S$,

$\begin{array}{c}{R}_S=2R+4R\\ =6R\end{array}$

Thus, the resistance across the circuit when resistors $R_2$ and $R_3$ are connected in series combination is $6R$.

The resistors $R_4$ and $R_S$ are connected in parallel combination as shown in figure 2.

Figure (2)

Formula to calculate the resistance when the resistors are connected in parallel is,

$\frac{1}{R_P}=\frac{1}{R_4}+\frac{1}{R_S}$ (2)

Here,

$R_P$ is the value of the resistance when the resistors are connected in parallel.

$R_4$ is the value of resistor 4.

Substitute $3R$ for $R_4$, $6R$ for $R_S$ in equation (2) to find $R_P$,

$\begin{array}{c}\frac{1}{R_P}=\frac{1}{3R}+\frac{1}{6R}\\ R_P=2R\end{array}$

Thus, the value of the resistance when the resistors are connected in parallel is $2R$.

The resistors $R_1$ and $R_P$ are connected in series combination as shown in figure 3.

Figure (3)

Formula to calculate the equivalent resistance across the circuit is,

$R_{\text{eq}}=R_1+R_P$ (3)

Here,

$R_{\text{eq}}$ is the equivalent resistance across the circuit.

$R_1$ is the value of resistor 1.

Substitute $R$ for $R_1$, $2R$ for $R_P$ in equation (3) to find $R_{\text{eq}}$,

$\begin{array}{c}{R}_{\text{eq}}=R+2R\\ =3R\end{array}$

Thus, the equivalent resistance across the circuit is $3R$.

The equivalent resistance is shown in the figure 4.

Figure (4)

Formula to calculate the current across the circuit is,

$I=\frac{V}{R_{\text{eq}}}$ (4)

Here,

$I$ is the current across the circuit.

$V$ is the voltage across the circuit.

Substitute $3R$ for $R_{\text{eq}}$, $\epsilon$ for $V$ in equation (4) to find $I$,

$I=\frac{\epsilon }{3R}$

Thus, the current across the circuit is $\frac{\epsilon }{3R}$.

Formula to calculate the voltage across the $R_P$ resistor is,

$V_{R_P}=I{R}_P$ (5)

Here,

$V_{R_P}$ is the voltage across the $R_P$ resistor.

Substitute $2R$ for $R_P$, $\frac{\epsilon }{3R}$ for $I$ in equation (5) to find $V_{R_P}$,

$\begin{array}{c}{V}_{R_P}=\left(\frac{\epsilon }{3R}\right)2R\\ =\frac{2\epsilon }{3}\end{array}$

Thus, the voltage across the $R_P$ resistor is $\frac{2\epsilon }{3}$.

Thus, the voltage across the $R_4$ resistor and $R_S$ resistor is $\frac{2\epsilon }{3}$.

Formula to calculate the voltage across the $R_1$ resistor is,

$V_{R_1}=I{R}_1$ (6)

Here,

$V_{R_1}$ is the voltage across the $R_1$ resistor.

Substitute $R$ for $R_1$, $\frac{\epsilon }{3R}$ for $I$ in equation (6) to find $V_{R_1}$,

$\begin{array}{c}{V}_{R_1}=\left(\frac{\epsilon }{3R}\right)R\\ =\frac{\epsilon }{3}\end{array}$

Thus, the voltage across the $R_1$ resistor is $\frac{\epsilon }{3}$.

Formula to calculate the current across the $R_S$ resistor is,

$I_{R_S}=\frac{V_{R_P}}{R_S}$ (7)

Here,

$I_{R_S}$ is the current across the $R_S$ resistor.

Substitute $6R$ for $R_S$, $\frac{2\epsilon }{3}$ for $V_{R_P}$ in equation (7) to find $I_{R_S}$,

$\begin{array}{c}{I}_{R_S}=\frac{\left(\frac{2\epsilon }{3}\right)}{6R}\\ =\frac{\epsilon }{9R}\end{array}$

Thus, the current across the $R_S$ resistor is $\frac{\epsilon }{9R}$.

Formula to calculate the current across the $R_4$ resistor is,

$I_{R_4}=\frac{V_{R_P}}{R_4}$ (8)

Here,

$I_{R_4}$ is the current across the $R_4$ resistor.

Substitute $3R$ for $R_4$, $\frac{2\epsilon }{3}$ for $V_{R_P}$ in equation (8) to find $I_{R_4}$,

$\begin{array}{c}{I}_{R_4}=\frac{\left(\frac{2\epsilon }{3}\right)}{3R}\\ =\frac{2\epsilon }{9R}\end{array}$

Thus, the current across the $R_4$ resistor is $\frac{2\epsilon }{9R}$.

Formula to calculate the voltage across the $R_2$ resistor is,

$V_{R_2}=I_{R_S}{R}_2$ (9)

Here,

$V_{R_2}$ is the voltage across the $R_2$ resistor.

Substitute $2R$ for $R_2$, $\frac{\epsilon }{9R}$ for $I_{R_S}$ in equation (9) to find $V_{R_2}$,

$\begin{array}{c}{V}_{R_2}=\left(\frac{\epsilon }{9R}\right)2R\\ =\frac{2\epsilon }{9}\end{array}$

Thus, the voltage across the $R_2$ resistor is $\frac{2\epsilon }{9}$.

Formula to calculate the voltage across the $R_3$ resistor is,

$V_{R_3}=I_{R_S}{R}_3$ (10)

Here,

$V_{R_3}$ is the voltage across the $R_3$ resistor.

Substitute $4R$ for $R_3$, $\frac{\epsilon }{9R}$ for $I_{R_S}$ in equation (10) to find $V_{R_3}$,

$\begin{array}{c}{V}_{R_3}=\left(\frac{\epsilon }{9R}\right)4R\\ =\frac{4\epsilon }{9}\end{array}$

Thus, the voltage across the $R_3$ resistor is $\frac{4\epsilon }{9}$.

Conclusion:

Therefore, the potential difference across $R_1$ resistor is $\frac{\epsilon }{3}$, potential difference across $R_2$ resistor is $\frac{2\epsilon }{9}$, potential difference across $R_3$ resistor is $\frac{4\epsilon }{9}$, potential difference across $R_4$ resistor is $\frac{2\epsilon }{3}$.

To determine

The current in each resistor in terms of $I$.

Answer to Problem 15P

The current across $R_1$ resistor is $I$, current across $R_2$ resistor is $\frac{I}{3}$, current across $R_3$ resistor is $\frac{I}{3}$, current across $R_4$ resistor is $\frac{2I}{3}$.

Explanation of Solution

Formula to calculate the value of $\epsilon$ is,

$V=\epsilon =I{R}_{\text{eq}}$ (11)

Substitute $3R$ for $R_{\text{eq}}$ in equation (11) to find $\epsilon$,

$\epsilon =3IR$

Thus, the value of $\epsilon$ is $3IR$.

Formula to calculate the current across the $R_1$ resistor is,

$I_{R_1}=\frac{V_{R_1}}{R_1}$ (12)

Here,

$I_{R_1}$ is the current across the $R_1$ resistor.

Substitute $R$ for $R_1$, $\frac{\epsilon }{3}$ for $V_{R_1}$ in equation (12) to find $I_{R_1}$,

$\begin{array}{c}{I}_{R_1}=\frac{\left(\frac{\epsilon }{3}\right)}{R}\\ =\frac{\epsilon }{3R}\end{array}$ (13)

Substitute $3IR$ for $\epsilon$ in equation (13) to find $I_{R_1}$,

$\begin{array}{c}{I}_{R_1}=\frac{3IR}{3R}\\ =I\end{array}$

Thus, the current across the $R_1$ resistor is $I$.

Formula to calculate the current across the $R_2$ resistor is,

$I_{R_2}=\frac{V_{R_2}}{R_2}$ (14)

Here,

$I_{R_2}$ is the current across the $R_2$ resistor.

Substitute $2R$ for $R_2$, $\frac{2\epsilon }{9}$ for $V_{R_2}$ in equation (14) to find $I_{R_2}$,

$\begin{array}{c}{I}_{R_2}=\frac{\left(\frac{2\epsilon }{9}\right)}{2R}\\ =\frac{\epsilon }{9R}\end{array}$ (15)

Substitute $3IR$ for $\epsilon$ in equation (15) to find $I_{R_2}$,

$\begin{array}{c}{I}_{R_2}=\frac{3IR}{9R}\\ =\frac{I}{3}\end{array}$

Thus, the current across the $R_2$ resistor is $\frac{I}{3}$.

Formula to calculate the current across the $R_3$ resistor is,

$I_{R_3}=\frac{V_{R_3}}{R_3}$ (16)

Here,

$I_{R_3}$ is the current across the $R_3$ resistor.

Substitute $4R$ for $R_3$, $\frac{4\epsilon }{9}$ for $V_{R_3}$ in equation (16) to find $I_{R_3}$,

$\begin{array}{c}{I}_{R_3}=\frac{\left(\frac{4\epsilon }{9}\right)}{4R}\\ =\frac{\epsilon }{9R}\end{array}$ (17)

Substitute $3IR$ for $\epsilon$ in equation (17) to find $I_{R_3}$,

$\begin{array}{c}{I}_{R_3}=\frac{3IR}{9R}\\ =\frac{I}{3}\end{array}$

Thus, the current across the $R_3$ resistor is $\frac{I}{3}$.

Formula to calculate the current across the $R_4$ resistor is,

$I_{R_4}=\frac{V_{R_4}}{R_4}$ (18)

Here,

$I_{R_4}$ is the current across the $R_4$ resistor.

Substitute $3R$ for $R_4$, $\frac{2\epsilon }{3}$ for $V_{R_4}$ in equation (18) to find $I_{R_4}$,

$\begin{array}{c}{I}_{R_4}=\frac{\left(\frac{2\epsilon }{3}\right)}{3R}\\ =\frac{2\epsilon }{9R}\end{array}$ (19)

Substitute $3IR$ for $\epsilon$ in equation (17) to find $I_{R_4}$,

$\begin{array}{c}{I}_{R_4}=\frac{2\left(3IR\right)}{9R}\\ =\frac{2I}{3}\end{array}$

Thus, the current across the $R_4$ resistor is $\frac{2I}{3}$.

Conclusion:

Therefore, the current across $R_1$ resistor is $I$, current across $R_2$ resistor is $\frac{I}{3}$, current across $R_3$ resistor is $\frac{I}{3}$, current across $R_4$ resistor is $\frac{2I}{3}$.

To determine

The effect in the current in each of the resistors, if $R_3$ is increased.

Answer to Problem 15P

The value of current across $R_1$ resistor increases and the value of current across $R_1,R_2\text{and}{R}_4$ resistor decreases.

Explanation of Solution

If the value of the $R_3$ is increased, then the value of current across $R_3$ resistor decreases as resistance is inversely proportional to the current.

Since, the current remains same in series combination. So, the value of current across $R_2$ resistor decreases.

If the value of the $R_3$ is increased, then the overall resistance in parallel combination decreases. The voltage remains same in parallel combination. As the voltage is directly proportional to the resistance. Thus, the voltage across the $R_4$ resistor decreases which results in increasing the current across the $R_4$ resistor.

Thus, the current across the $R_1$ resistor decreases as the voltage across the $R_1$ resistor increases.

Conclusion:

Therefore, the value of current across $R_1$ resistor increases and the value of current across $R_1,R_2\text{and}{R}_4$ resistor decreases.

To determine

The new values of the current in each resistor in terms of $I$ and the original current in the battery.

Answer to Problem 15P

The current across $R_1$ resistor is $\frac{3I}{4}$, current across $R_2$ resistor is $0$, current across $R_3$ resistor is $0$, current across $R_4$ resistor is $\frac{3I}{4}$ and the original current in the battery is $\frac{3I}{4}$.

Explanation of Solution

If $R_3$ tends to infinity then, current across $R_2$ resistor and $R_3$ resistor is $0$.

${I^{\prime }}_{R_2}={I^{\prime }}_{R_3}=0$

Here,

${I^{\prime }}_{R_2}$ is the new value of current across $R_2$ resistor.

${I^{\prime }}_{R_3}$ is the new value of current across $R_3$ resistor.

The resistors $R_1$ and $R_4$ are connected in series combination.

Formula to calculate the resistance across the circuit when resistors $R_1$ and $R_4$ are connected in series combination.

${R^{\prime }}_S=R_1+R_4$ (20)

Here,

${R^{\prime }}_S$ is the resistance across the circuit when resistors $R_1$ and $R_4$ are connected in series combination.

$R_1$ is the value of resistor 1.

$R_4$ is the value of resistor 4.

Substitute $R$ for $R_1$, $3R$ for $R_4$ in equation (20) to find ${R^{\prime }}_S$,

$\begin{array}{c}{R^{\prime }}_S=R+3R\\ =4R\end{array}$

Thus, the resistance across the circuit when resistors $R_1$ and $R_4$ are connected in series combination is $4R$.

From equation (11), the value of $\epsilon$ is $3IR$.

From equation (12), formula to calculate the current across the $R_1$ resistor when $R_3$ tends to infinity is,

${I^{\prime }}_{R_1}=\frac{\epsilon }{{R^{\prime }}_S}$ (21)

Here,

${I^{\prime }}_{R_1}$ is the new value of current across $R_1$ resistor.

Substitute $3IR$ for $\epsilon$, $4R$ for ${R^{\prime }}_S$ in equation (21) to find ${I^{\prime }}_{R_1}$,

$\begin{array}{c}{I^{\prime }}_{R_1}=\frac{3IR}{4R}\\ =\frac{3I}{4}\end{array}$

Thus, the current across the $R_1$ resistor when $R_3$ tends to infinity is $\frac{3I}{4}$.

As the resistors $R_1$ and $R_4$ are connected in series combination so the value of current across $R_1$ resistor is equal to the current across $R_4$ resistor.

${I^{\prime }}_{R_1}={I^{\prime }}_{R_4}=\frac{3I}{4}$

Here,

${I^{\prime }}_{R_4}$ is the new value of current across $R_4$ resistor.

Thus, the original current in the battery is $\frac{3I}{4}$.

Conclusion:

Therefore, the current across $R_1$ resistor is $\frac{3I}{4}$, current across $R_2$ resistor is $0$, current across $R_3$ resistor is $0$, current across $R_4$ resistor is $\frac{3I}{4}$ and the original current in the battery is $\frac{3I}{4}$.