DC Circuits - Problem Solving

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:

\(V = I R\)

\(R=V/I\)

\(R = 3.0 \Omega\)

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.

\(V = V_1 +V_2 +V_3 +V_4\)

\(V = 5.0+8.0+7.0+10 = 30 V\)

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 \( I = 12.0A\).

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 \(R = 6.7 \Omega\).

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

\(\frac {1}{R_\mathrm {equivalent}} = \sum \frac {1}{R_i}\)

we can find the equivalent resistance of these branches.