# DC Circuits - Problem Solving

## Example Problem on Ohm's Law: The Basic Circuit

### Question

An emf source of $6.0V$ is connected to a purely resistive lamp and a current of $2.0$ amperes flows. All the wires are resistance-free. What is the resistance of the lamp?

### Hints

1. Where in the circuit does the gain in potential energy occur?
2. Where in the circuit does the loss of potential energy occur?
3. What is Ohm's Law?

#### Solution

The gain of potential energy occurs as a charge passes through the battery, that is, it gains a potential of $\varepsilon = 6.0V$. No energy is lost to the wires, since they are assumed to be resistance-free. By conservation of energy, the potential that was gained (i.e. $\varepsilon= V = 6.0V$) must be lost in the resistor. So, by Ohm's Law:

$V = I R$

$R=V/I$

$R = 3.0 \Omega$

## Example of Problem on Resistors in Series

### Question

The current flowing in a circuit containing four resistors connected in series is $I = 1.0 A$. The potential drops across the first, second and third resistors are, respectively: $V = 5 V$, $V = 8 V$ and $V = 7 V$.
The equivalent resistance of the circuit is $R = 30 \Omega$.

Find the total voltage supplied by the battery, and also current, voltage drop, and resistance of each resistor in the circuit.

### Hints

1. How are resistors related when connected in series?
2. What is true about potential drops of resistors when connected in series?
3. You will need to use Ohm's Law.

#### Solution

First, let's label the diagram with the information given in the question.

There are several ways of solving this problem (see alternate solutions), but this tutorial will only go through one of these ways.

Because the resistors are connected in series, then the same current flows through each one. Using the Ohm's Law, we can find the resistances of the first, second and third resistors.

$R_1 = \frac {V_1}{I}, R_2 = \frac{V_2}{I} , R_3 = \frac {V_3}{I}$

$R_1 = \frac {5.0}{1.0} = 5.0 \quad \Omega R_2 = \frac{8.0}{1.0} = 8.0 \quad \Omega R_3 = \frac {7.0}{1.0} = 7.0\Omega$

Now, using the equivalent resistance, we can find the resistance in the fourth resistor. This is a series circuit, so the equivalent resistance is the sum of the individual resistances.

$R_ \mathrm {equivalent} = R_1 + R_2 +R_3+R_4$

$R_4 = R _\mathrm {equivalent} - ( R_1+R_2+R_3)$

$R_4 = 30 - (5.0 + 8.0 + 7.0) = 10 \Omega$

The current flowing through the fourth resistor is also $I=1.0A$. Using Ohm's Law again, we find the voltage across this resistor.

$V_4 = I \cdot R_4$

$V_4 = (1.0)\cdot (10) = 10V$

The total voltage supplied by the battery must equal to the total voltage drop across the circuit (this is known as Kirchhoff's Voltage Law). So, we must sum up the voltage drops across the resistors.

$V = V_1 +V_2 +V_3 +V_4$

$V = 5.0+8.0+7.0+10 = 30 V$

## Example Problem on Resistors in Parallel

### Question

In the following schematic diagram, find the total current, I.

### Hints

1. You will need Ohm's Law.
2. How are resistors related when connected in parallel?
3. What is the potential drop across each resistor?
4. How does current behave in parallel branches?

#### Solution

We know the total potential of this circuit,

$\varepsilon = 12.0 V$

So, between points $A$ and $B$, the potential must drop $12.0V$. Also, the potential drop across branches of a circuit are equal. That is,

$V_1 = V_2 = V_3 = \varepsilon = 12.0V$

We can use Ohm's Law

$V = IR$

or

$I = V/R$

to find the current across each resistor.

$I_1 = \frac {V_1}{R_1} = \frac {12.0V}{2.0 \Omega} = 6.0A$

$I_2 = \frac{V_2}{R_2} = \frac{12.0V}{3.0 \Omega} = 4.0A$

$I_3 = \frac {V_3}{R_3} = \frac {12.0V}{6.0 \Omega} = 2.0A$

Recall that the currents through branches of a parallel circuit add to give the total current. That is, the total current 'splits up' so that part of the total current travels down each branch. Because of conservation of charge, the sum of the currents in each branch must equal the amount going into the branch. (This is Kirchhoff's Current Law.)

So, adding up the three currents, we get:

$I = I_1 +I_2 + I_3$

$= 6.0 + 4.0 +2.0 = 12.0 A$

So, the total current is $I = 12.0A$.

## Example Problems on Resistors in Combination Circuits

### Hints

1. Which resistors are in parallel and which are in series?
2. Is this circuit composed of small groups of parallel resistors, all connected in series? Or is it composed of groups of series resistors, connected in parallel?

#### Solution

This circuit is composed of 3 'elements' connected in series: the group of parallel resistors between $A$ and $B$, the single resistor $R_3$, and the group of parallel resistors between $C$ and $D$.

First, we will find the equivalent resistance between $A$ and $B$.

Here, we have two resistors, $R_1$ and $R_2$, connected in parallel. Using the formula for resistors connected in parallel:

$\frac {1}{R_\mathrm{equivalent}} = \sum \frac{1}{R_i}$

we can find the equivalent resistance between points $A$ and $B$. Let's call this equivalent resistance $R_{AB}$.

$\frac {1}{R_{AB}} = \frac {1}{R_1} + \frac {1}{R_2}$

$= \frac {1}{10.0}+ \frac{1}{4.0}$

$= 0.35$

$R_{AB} = 2.857 \Omega$

Now, we'll find the equivalent resistance between $C$ and $D$, and will call it $R_{CD}$. Using the equation from above for resistors connected in parallel,

$\frac {1}{R_{CD}} = \frac {1}{R_4} + \frac {1}{R_5}$

$\frac {1}{R_{CD}} = \frac{1}{8.0}+\frac{1}{1.0}$

$\frac {1}{R_{CD}} = 1.125$

$R_{CD} = 0.889 \Omega$

Replacing the two parallel sections with their equivalent resistances, and redrawing the circuit, we get the circuit in Figure 2. We see that there are three resistances connected in series: $R_{AB}$, $R_3$, and $R_{CD}$. Using the formula for resistors in series,

we can find the equivalent resistance of the circuit.

So the equivalent resistance of this circuit is $R = 6.7 \Omega$.

### Question 2

Figure 7 shows part of a circuit. It consists of resistors combined in both parallel and series configurations. Find the equivalent resistance.

### Hints

1. What is the equivalent resistance for resistors in parallel?
2. In series?

#### Solution

In this partial circuit, there are three main branches, branch $AB$, branch $CD$, branch $EF$. As you can see, branch $AB$ contains two resistors in series, $R_1$ and $R_2$. Branch $CD$ has just one resistor, $R_3$. Finally, there are two resistors in branch $EF$.

Let's look at branch $AB$ first. We will simplify this branch by finding the equivalent resistance between $A$ and $B$. Note that $R_1$ is connected in series with $R_2$. Using the equation for resistors in series

$R_\mathrm {equivalent} = \sum R_i$

we can find $R_{AB}$.

$R_{AB} = R_1 + R_2$

$R_{AB} = 1.0 + 2.0$

$R_{AB} = 3.0 \Omega$

Now, in branch $CD$ there is only one resistor, so this branch cannot be simplified further.

In branch $EF$, however, there are two resistors, connected in series with one another. Using the equation for resistors in series, we can find the equivalent resistance in branch $EF$, $R_{EF}$.

$R_{EF} = R_4 + R_5$

$R_{EF} = 4.0 + 5.0$

$R_{EF} = 9.0$

We can redraw Circuit 2 using $R_{AB}$, $R_3$, and $R_{EF}$, as seen in Figure 8. This circuit has been simplified to a parallel circuit, with three resistances in parallel. Using the formula for resistors connected in parallel
$\frac {1}{R_\mathrm {equivalent}} = \sum \frac {1}{R_i}$
$\frac {1}{R} = \frac {1}{R_{AB}} + \frac {1}{R_3}+ \frac{1}{R_{EF}}$
$\frac{1}{R} = \frac{1}{3.0} + \frac{1}{3.0}+ \frac{1}{9.0}$
$R = 1.286 \Omega$