Finding the voltage across each resistor

I’ve just started an engineering course, and my 1st module is electrical engineering. Unfortunately my teacher is a 100 years old and hasn’t explained Kirchoff’s law to us in an understanding manor. In this question I’ve been asked to find the voltage across each resistor. Please could someone explain to me how this is done. The simpler the explanation the better, I honestly won’t find it patronising, it would be a great help. Thanks in advance

  • \$\begingroup\$ No textbook for the course ? \$\endgroup\$ – efox29 Jan 22 '16 at 16:46
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    \$\begingroup\$ Do you know how to solve the circuit if there was only voltage source present ? \$\endgroup\$ – efox29 Jan 22 '16 at 16:47
  • \$\begingroup\$ no i wouldn't know \$\endgroup\$ – Ben Jan 22 '16 at 17:18
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    \$\begingroup\$ Perhaps the teacher is excellent, and you are the opposite. \$\endgroup\$ – Chu Jan 22 '16 at 20:11
  • \$\begingroup\$ FYI there are various on-line lectures that are more suitable than this site when you know nothing. Getting a nutshell response here won't be a good a substitute for that. Youtube is your friend Reasonable examples: youtube.com/… and ocw.mit.edu/courses/electrical-engineering-and-computer-science/… Good luck. \$\endgroup\$ – Fizz Jan 22 '16 at 20:53

Kirchoff's laws don't all need to be used to solve one circuit. You can go with Kirchoff's nodal analysis (KCL), or Kirchoff's Voltage law (KVL)

Kirchoff's nodal analysis (KCL):

Sum of the currents going into a node equals the sum of the currents coming out.

So.. Let us assign a variable name to (i.e. Vx, Va) to any connection with more than 2 ends. So assign Vx to that node on the top. Then you need to characterize the algebra problem by defining each current and how they equal relative to each other using this law.

Then try to analyze the currents across each branch with respect to Va and to ground or voltage source that is in series.

Kirchoff's Voltage law (KVL):

Sum of voltages in a loop sum up to zero.

A loop in this context is a series of circuit components whose path starts at a source path and leads down back to the that source path, path = < c1, c2, c3, c1 > (where cx = component x) (i.e. 10V, 5Ohm, 10Ohm, back to 10V)

So.. assign all existing current loops, i.e. left loop can be Current 1, I1 right loop can be current 2, I2 and the current going down between the two (current 3, I3) is a mix of the two, which relatively depends on the direction assigned to each current (I1, and I2)

Then you need to characterize the algebra problem to find the voltage across each component in each loop.


It's too bad that you do not have a great teacher. For me, I find that Kirchoff's laws are just amazing. And they really are quite simple. Just by using Ohm's Law (\$v = i \times R\$), and Kirchoff's laws you can solve any network that consists of passive components.

Kirchoff's Current Law

The easiest one to understand is the current law. Think about a "cut" through a wire. Since no current flows out of the wire into air, the amount of current into the cut, must be the same as the current out of the cut area. The same idea applies to a junction, so $$ i_\text{in} = i_\text{out}, $$ simple as that.

  • A note about directions. Make a habit out of writing arrows for your currents. It doesn't matter which way you write them, because the cool thing about the equations is that if you guessed the wrong direction, the resulting current simply gets negative.
  • You may also have seen the law as $$ i_1 + \cdots + i_n = 0. $$ This assumes that the direction is either all into or out of the junction (the node). So, for instance for the equation above, \$i_\text{in} - i_\text{out} = 0\$ makes sense, since the current going into the cut on the output side must be the negated current out of the cut.
  • Voltage levels in a circuit are relative. In order for you to measure voltage, you need two leads. To make things simpler, we usually define a specific node which we consider 0V, which we call ground. It usually makes sense to attribute ground to the negative battery connection, because then most other nodes will be positive (compared to ground).

Using KCL

I will solve your problem using KCL and Ohm's Law. You have two junctions, the top one and the bottom one. I will call the bottom one ground, since it is connected to both negative battery terminals. So we define the voltage level there to be 0V.

There are two other nodes, the connection between the left battery (\$+\$-terminal) and the \$5\Omega\$ resistor, and the right battery and the \$6\Omega\$ resistor. We know the voltages of those because of the batteries (10 V and 12 V).

So, we are left with a single voltage, the top junction, which I will call \$V_x\$.

We see that the voltage drop from the left battery over the left resistor is \$10 - V_x\$. Using Ohm's Law: $$ (10\;\text{V}) - V_x = i_\text{left} \times (5\;\Omega) $$ This means (remember, we want currents for KCL): $$ i_\text{left} = \frac{10 - V_x}{5} $$ Similarly, for the right side, we get $$ i_\text{right} = \frac{12 - V_x}{6} $$ For the middle resistor, we have a voltage drop of \$V_x - V_\text{ground} = V_x - 0 = V_x\$, so defining \$i_\text{middle}\$ as downwards: $$ i_\text{middle} = \frac{V_x}{10} $$

Now we have all what we need. KCL gives us $$ i_\text{left} + i_\text{right} - i_\text{middle} = 0 $$ where you notice the negation due to my direction of the middle resistor. Plugging in the currents from above: $$ \frac{10 - V_x}{5} + \frac{12 - V_x}{6} - \frac{V_x}{10} = 0 $$ Getting rid of fractions by multiplying with 30: $$ 60 - 6 V_x + 60 - 5 V_x - 3V_x = 0 \\ 120 - 14 V_x = 0 \\ 120 = 14 V_x \\ V_x = \frac{120}{14} = 8\frac{4}{7}\approx 8.57 $$ So the voltage at the upper junction (compared to ground) is 8.57 V.


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