If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Electromagnetism (Essentials) - Class 12th>Unit 6

Lesson 4: Simplifying resistor networks

# Example: Analyzing a more complex resistor circuit

Master the art of solving complex circuit problems with this guide. Learn how to simplify circuits by finding equivalent resistances, understand the concept of resistors in series and parallel, and apply Ohm's law to find the current. Created by Sal Khan.

## Want to join the conversation?

• Can we calculate the current in each of the resistors ?
• its quite easy infact if you just remember these two basic concepts....
1). current is same across all resistors in series..
2). to find current in a particular resistance (r1) connected in parallel to others (equivalent resistance = R.Eq )...

current through r1 = * [ eq. resistance of all (connected in parallel )except r1 ] X {total current flowing through " R.Eq " ] / [ R.Eq ] *

P.S. - derivation through V = IR or krihoff voltage law..
`code`
``if (x < 0) {    return;}``
(1 vote)
• Why in a real life application would we use multiple resistors in both series and parallel configurations together?
• How can we calculate the internal resistance inside a battery?
• You can't calculate it, but you can measure it by measuring how the terminal voltage changes as the current changes, and then applying ohm's law.
• what i think here is that when the circuit is in parallel the equivalent resistance is less than the least resistance in the parallel circuit .. and when in series... the equi. resistance is greater than the largest resistance in the series circuit.. am i right??
• You are absolutely correct. The series circuit, I need not answer. The parallel circuit, I will prove it intuitively. A wire is also a resistence, and a good example. When we see the electrons bumping into each other and into the atomic nuclei inside the wire, we call that property resistence. Kind of like viscosity in fluids. So if a wire gets longer, it has more chances to bump into something. Also, if a wire gets thicker, an electron will have more chance to AVOID collisions. So a thicker wire means less resistence. In analogy, if one has few wires in parallel, it will have same effects as making one wire thicker. So multiple wires would mean less resistence than the least of one
• If you call these complex, Idk what the hell I'm solving
• can u please make it clear how to determine the direction of the current flow then it would be helpful
• in a simple DC circuit, the direction of current is from positive terminal to negative
• what is e.m.f and terminal voltage of a cell?how are they related to internal resistance?
(1 vote)
• EMF is electromotive force, and it's another way to refer to voltage.

Terminal voltage is the voltage that the battery can deliver into a working circuit. If you have a battery, you might think of it as say a 9 volt battery, but that 9v is what it can deliver when it is not having to put out much current - in other words, when the resistance in the rest of the circuit is very high.

When you connect a batter to a circuit where the resistance is low, then you have a situation where the battery is trying to put out a lot of current. If you want to calculate how much current there will actually be, now you can't ignore the resistance that is internal to the battery.

To make the math easy, let's say we have a 10v battery connected to a 10k ohm resistor. The current is going to be very small, only 1 mA, but technically that is only an approximation, because the battery has resistance, too. So, say the battery has internal resistance of 1 ohm. Now the real resistance of the circuit is 10.001k ohm, and you can see, that's not going to make any meaningful difference when you calculate the current. It's still really, really close to 1 mA. The voltage drop throughout the circuit is going to occur almost entirely across the 10k resistor. The battery terminal voltage therefore will be really close to 10 V.

Now, connect the battery instead to a resistor that is only 10 ohm. If you had a batter with no internal resistance, you have 10 V/ 10 ohm = 1 amp. But now you can see that the internal resistance matters. The real resistance around the circuit is not 10, but 11. So the current is closer to 0.9A (I am rounding 10/11) And now you can see that the voltage drop that occurs across the resistor will be 0.9A x 10 ohm = 9V, and that's the terminal voltage of the battery. Where is the other 1 V we expected to see, since the voltage all the way around the loop is supposed to sum to 0 (+10 from battery, -9 lost across resistor, 1 more volt needs to be accounted for). That voltage drop occurs across the internal resistance of the battery itself.
• The circuit in the video was 2D and therefore could be analyzed step by step. However what would happen if resistors were connected in a 3D circuit. For example, a circuit is made in the shape of a cube where there is a resistor of 1 ohm on each edge of the cube. In this case, what would be the total resistance.
• how do you solve for the current through each individual resistor in the circuit?
(1 vote)
• If it is a series circuit, the current is the same through all the resistors. Find the total resistance, divide the voltage by that, and you will have the current.

If it is a parallel circuit, then just do V/R for each of the resistors, and you will have the currents.