Chapter 21, Problem 40P

Four resistors are connected to a battery as shown in Figure P21.40. The current in the battery is I, the battery emf is ε, and the resistor values are R₁ = R, R₂ = 2R, R₃ = 4R, and R₄ = 3R. (a) Rank the resistors according to the potential difference across them, from largest to smallest. Note any cases of equal potential differences. (b) Determine the potential difference across each resistor in terms of ε. (c) Rank the resistors according to the current in them, from largest to smallest. Note any cases of equal currents. (d) Determine the current in each resistor in terms of I. (e) If R₃ is increased, what happens to the current in each of the resistors? (f) 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 P21.40

To determine

The rank of the resistors according to the potential difference across them from largest to smallest.

Answer to Problem 40P

The rank the resistors according to the potential difference across them from largest to smallest is $\underset{\_}{\Delta V_4>\Delta V_3>\Delta V_1>\Delta V_2}$.

Explanation of Solution

The resistor $R_2$ and $R_3$ are in series and that combination is parallel to $R_4$.

Write the equivalent resistance for the combination of resistors $R_2$, $R_3$ and $R_4$.

    $\begin{array}{c}{R}_{\text{eq}}=\frac{R_4\left(R_3+R_2\right)}{R_4+\left(R_3+R_2\right)}\\ =\frac{3R\left(4R+2R\right)}{3R+\left(4R+2R\right)}\\ =2R\end{array}$        (I)

Here, $R_{\text{eq}}$ is the equivalent resistance and $R$ is the resistance.

The $R_{\text{eq}}$ is in series with resistor $R_1$.

Write the resistance of the circuit.

    $\begin{array}{c}{R}_T=R_1+R_{\text{eq}}\\ =R+2R\\ =3R\end{array}$        (II)

Here, $R_T$ is the total resistance of the circuit.

In series combinations, the potential is dropped in proportion to the resistance.

Resistor $R_1$ gets $\frac{1}{3}$ of the battery voltage and $R_2,R_3\text{and}{R}_4$ are in parallel combination get $\frac{2}{3}$ of the battery voltage. The voltage across the $R_4$ is $\frac{2}{3}$ of the battery voltage, $R_2$ and $R_3$ must share this voltage. $\frac{1}{3}$ and $\frac{2}{3}$ of the battery voltages are shared across $R_2$ and $R_3$.

Conclusion:

Therefore, the rank of resistors according to the potential difference across them from largest to smallest is $\underset{\_}{\Delta V_4>\Delta V_3>\Delta V_1>\Delta V_2}$.

To determine

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

Answer to Problem 40P

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

Explanation of Solution

In series combinations, the potential is dropped in proportion to the resistance.

Resistor $R_1$ gets $\frac{1}{3}$ of the battery voltage and $R_2,R_3\text{and}{R}_4$ are in parallel combination get $\frac{2}{3}$ of the battery voltage. The voltage across  $R_4$ is $\frac{2}{3}$ of the battery voltage, $R_2$ and $R_3$ must share this voltage. $\frac{1}{3}$ and $\frac{2}{3}$ of the battery voltages are shared across $R_2$ and $R_3$.

Conclusion:

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

To determine

The rank of the resistors according to the current in them from largest to smallest.

Answer to Problem 40P

The rank of the resistors according to the current in them from largest to smallest is $\underset{\_}{I_1>I_4>I_2=I_3}$.

Explanation of Solution

All the current will pass through the resistor $R_1$ since it is series to the battery. Then the current splits at the parallel combination. The resistor $R_4$ gets more current because the resistance in that branch is less than other branch which is the series combination of $R_2$ and $R_3$.

The ranking of the resistors according to the current in them from largest to smallest is $I_1>I_4>I_2=I_3$.

Conclusion:

Therefore, the rank resistors according to the current in them from largest to smallest is $\underset{\_}{I_1>I_4>I_2=I_3}$.

To determine

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

Answer to Problem 40P

The current in $R_1$ is $\underset{\_}{I}$, $R_2$ is $\underset{\_}{\frac{I}{3}}$, $R_3$ is $\underset{\_}{\frac{I}{3}}$ and $R_4$ is $\underset{\_}{\frac{2I}{3}}$.

Explanation of Solution

The resistor $R_1$ is in series so the current through $R_1$ is $I$. Since the resistor $R_2$ and $R_3$ are connected in series, the current through these resistors is same. The resistance of $R_2$ and $R_3$ in series is twice that of resistor $R_4$, twice as much current goes through $R_4$ than through $R_2$ and $R_3$.

As a result, the current in $R_1$ is $I$, $R_2$ is $\frac{I}{3}$, $R_3$ is $\frac{I}{3}$ and $R_4$ is $\frac{2I}{3}$.

Conclusion:

Therefore, the current in $R_1$ is $\underset{\_}{I}$, $R_2$ is $\underset{\_}{\frac{I}{3}}$, $R_3$ is $\underset{\_}{\frac{I}{3}}$ and $R_4$ is $\underset{\_}{\frac{2I}{3}}$.

To determine

If $R_3$ is increased, the current in each of the resistors.

Answer to Problem 40P

The value of current $I_4$ will increase and the current $I_1,I_2\text{and}{I}_3$ will decrease.

Explanation of Solution

As the value of resistance $R_3$ increases, simultaneously the resistance of the entire circuit will increase. The current through the resistor $R_1$ will decrease since the current in the circuit is same as the current through $R_1$. This decreases the potential difference across the resistor $R_1$ and increasing the potential difference across the parallel combination. The current through the $R_4$ will increased with the increased potential difference across the parallel combination. As a result, the value of current $I_4$ will increase and the current $I_1,I_2\text{and}{I}_3$ will decrease.

Conclusion:

Therefore, the value of current $I_4$ will increase and the current $I_1,I_2\text{and}{I}_3$ will decrease.

To determine

If $R_3\to \infty$, the new values of current in each resistor in terms of $I$.

Answer to Problem 40P

The current $I_1$ is $\underset{\_}{\frac{3I}{4}}$, $I_2=I_3=\underset{\_}{0}$ and $I_4$ is $\underset{\_}{\frac{3I}{4}}$.

Explanation of Solution

If the $R_3$ has an infinite resistance, the entire circuit can be considered with two resistors $R_1$ and $R_4$ in series with a battery. Because the infinite resistance of $R_3$ ceases any current passing through that branch. The equivalent resistance of the circuit is $4R$ and the current drops to $\frac{3}{4}$ of its original value due to the increased value of resistance by a factor of $\frac{4}{3}$. All this current will pass through $R_1$ and $R_4$. No current will flow through $R_2$ and $R_3$.

Conclusion:

Therefore, the current $I_1$ is $\underset{\_}{\frac{3I}{4}}$, $I_2=I_3=\underset{\_}{0}$ and $I_4$ is $\underset{\_}{\frac{3I}{4}}$.