Resistor Networks Further Revised

Two ohmic resistors (\(R_{1}\) and \(R_{2}\)) are connected in series with a cell with negligible internal resistance. Find the resistance of \(R_{2}\), given that the current flowing through \(R_{1}\) and \(R_{2}\) is \(\text{0.25}\) \(\text{A}\) and that the potential difference across the cell is \(\text{1.50}\) \(\text{V}\). \(R_{1}\) =\(\text{1.00}\) \(\text{Ω}\).

Step 1: Draw the circuit and fill in all known values.

Step 2: Determine how to approach the problem.

We can use Ohm's Law to find the total resistance R in the circuit, and then calculate the unknown resistance using:

\[R = R_{1} + R_{2}\]

because \(R_{1}\) and \(R_{2}\) are connected in series.

Step 3: Find the total resistance

\begin{align*} R & = \frac{V}{I} \\ & = \frac{\text{1.5}}{\text{0.25}} \\ & = \text{6}\text{ Ω} \end{align*}

Step 4: Find the unknown resistance

We know that:

\[R = \text{6.00}\text{ Ω}\]

and that

\[R_{1} = \text{1.00}\text{ Ω}\]

Since

\[R = R_{1} + R_{2}\]\[R_{2} = R - R_{1}\]

Therefore,

\[R_{1} = \text{5.00}\text{ Ω}\]

In the case of the circuit shown, calculate:

1. the potential difference of \(V_1\), \(V_2\) and \(V_3\) across the resistors \(R_1\), \(R_2\), and \(R_3\)

2. the resistance of \(R_3\).

Step 1: Determine how to approach the problem

We are given the potential difference across the cell and the current in the circuit, as well as the resistances of two of the three resistors. We can use Ohm's Law to calculate the potential difference across the known resistors. Since the resistors are in a series circuit the potential difference is \(V = V_1 + V_2 + V_3\) and we can calculate \(V_3\). Now we can use this information to find the potential difference across the unknown resistor \(R_3\).

Step 2: Calculate potential difference across \(R_1\)

Using Ohm's Law:\begin{align*} R_1 &= \frac{V_1}{I} \\ I \cdot R_1 &= I \cdot \frac{V_1}{I} \\ V_1 &= {I}\cdot{R_1}\\ &= \text{2} \cdot \text{2} \\ V_1 &= \text{4.00}\text{ V} \end{align*}

Step 3: Calculate potential difference across \(R_2\)

Use Ohm's Law:\begin{align*} R_2 &= \frac{V_2}{I} \\ I \cdot R_2 &= I \cdot \frac{V_2}{I} \\ V_2 &= {I}\cdot{R_2}\\ &= \text{2} \cdot \text{6} \\ V_2 &= \text{12.00}\text{ V} \end{align*}

Step 4: Calculate potential difference across \(R_3\)

Since the potential difference across all the resistors combined must be the same as the potential difference across the cell in a series circuit, we can find \(V_3\) using:\begin{align*} V &= V_1 + V_2 + V_3\\ V_3 &= V - V_1 - V_2 \\ &= \text{36} - \text{4} - \text{12} \\ V_3&= \text{20.00}\text{ V} \end{align*}

Step 5: Find the resistance of \(R_3\)

We know the potential difference across \(R_3\) and the current through it, so we can use Ohm's Law to calculate the value for the resistance:\begin{align*} R_3 &= \frac{V_3}{I}\\ &= \frac{ \text{20}}{ \text{2}} \\R_3&= \text{10.00}~\Omega\end{align*}

Step 6: Write the final answer

• \(V_1 = \text{4.00}\text{ V}\)
• \(V_2 = \text{12.00}\text{ V}\)
• \(V_3 = \text{20.00}\text{ V}\)
• \(R_3 = \text{10.00}~\Omega\)

Calculate the current (\(I\)) in this circuit if the resistors are both ohmic in nature.

Step 1: Determine what is required

We are required to calculate the total current flowing in the circuit.

Step 2: Determine how to approach the problem

Since the resistors are ohmic in nature, we can use Ohm's Law. However, there are two resistors in the circuit and we need to find the total resistance.

Step 3: Find the equivalent resistance in the circuit

Since the resistors are connected in parallel, the total (equivalent) resistance R is:

\[\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}.\]\begin{align*} \frac{1}{R} &= \frac{\text{1}}{R_1} + \frac{\text{1}}{R_2} \\ &= \frac{\text{1}}{\text{2}} + \frac{\text{1}}{\text{4}} \\ &= \frac{\text{2}+\text{1}}{\text{4}} \\ &= \frac{\text{3}}{\text{4}} \\ \text{Therefore. } R &= \frac{\text{4}}{\text{3}} = \text{1.33}~\Omega\end{align*}

Step 4: Apply Ohm's Law

\begin{align*} R&= \frac{V}{I} \\ R \cdot \frac{I}{R} &= \frac{V}{I} \cdot \frac{I}{R} \\ I &= \frac{V}{R}\\ I &= V \cdot \frac{\text{1}}{R}\\ &= (\text{12}) \left(\frac{\text{3}}{\text{4}}\right) \\ &= \text{9.00}\text{ A}\end{align*}

Step 5: Write the final answer

The total current flowing in the circuit is \(\text{9.00}\) \(\text{A}\).

An \(\text{18.00}\) \(\text{V}\) cell is connected to two parallel resistors of \(\text{2.00}\) \(\Omega\) and \(\text{6.00}\) \(\Omega\) respectively. Calculate the current through each of the ammeters when the switch is closed and when it is open.

Step 1: Determine how to approach the problem

We need to determine the current through the cell and each of the parallel resistors. We have been given the potential difference across the cell and the resistances of the resistors, so we can use Ohm's Law to calculate the current.

There are two alternative approaches we could adopt:

• we could use the fact that the potential difference across each of the resistors is the same as thepotential difference across the battery because they are in a parallel configuration and then useOhm's Law; or
• we could determine the equivalentresistance of the circuit and the total current and then use that to determine the currentthrough each of the resistors.

Step 2: Now determine the current through one of the parallel resistors

We know that for a configuration with just two resistors in parallel and a cell as in this case, the potential difference across the cell is the same as the potential difference across each of the resistors in parallel. For this circuit:\begin{align*}V &= V_1 = V_2 = \text{18.00}\text{ V}\end{align*}Let's start with calculating the current through \(R_1\) using Ohm's Law:\begin{align*}R_1 &= \frac{V_1}{I_1} \\I_1 &= \frac{V_1}{R_1} \\&= \frac{\text{18.00}}{\text{2.00}} \\I_1 &= \text{9.00}\text{ A}\end{align*}

Step 3: Calculate the current through the other parallel resistor

We can use Ohm's Law again to find the current in \(R_2\):\begin{align*}R_2 &= \frac{V_2}{I_2} \\I_2 &= \frac{V_2}{R_2} \\&= \frac{\text{18.00}}{\text{6.00}} \\I_2 &= \text{3.00}\text{ A}\end{align*}

Step 4: Calculate the total current

The current through each of the parallel resistors must add up to the total current through the cell:\begin{align*}I &= I_1 + I_2 \\&= \text{9.00} + \text{3.00} \\I_2 &= \text{12.00}\text{ A}\end{align*}

Step 5: When the switch is open

The branch through \(R_1\) is not complete so no current can flow through it.This means we can ignore it completely and consider a simple circuit with a battery and asingle resistor, \(R_2\), in it.

We can use Ohm's Law again to find the current in \(R_2\):\begin{align*}R_2 &= \frac{V_2}{I_2} \\I_2 &= \frac{V_2}{R_2} \\&= \frac{\text{18.00}}{\text{6.00}} \\I_2 &= \text{3.00}\text{ A}\end{align*}

Step 6: Write the final answer

When the switch is closed:

• The current through the cell is \(\text{12.00}\) \(\text{A}\).
• The current through the \(\text{2.00}\) \(\Omega\) resistor is \(\text{9.00}\) \(\text{A}\).
• The current through the \(\text{6.00}\) \(\Omega\) resistor is \(\text{3.00}\) \(\text{A}\).

When the switch is open:

• The current through the \(\text{6.00}\) \(\Omega\) resistor is \(\text{3.00}\) \(\text{A}\).