00:01Hi friends, learning is a continuous process.
00:05You won't become smart just by learning once.
00:10Two resistors are connected in parallel.
00:15Now, we are asked to calculate the current value through each resistor.
00:23Let's analyze this circuit in more detail.
00:28Assume the current entering the circuit is I, and the resistors are R1 and R2.
00:37The current is divided into I1 and I2.
00:43This circuit consists of a loop and two branching points.
00:51Since no current accumulates at each branching point or resistor, the algebraic sum of the currents at each branching point
00:58is zero.
01:01This is Kirchhoff's current law, I minus I1 minus I2 equals zero.
01:09From the voltage perspective, we can apply Kirchhoff's voltage law.
01:18Assume the loop rotates counterclockwise.
01:25There are two electrical components.
01:29For the top resistor, the voltage sign is positive because the loop and current are in opposite directions.
01:37For the bottom resistor, the voltage sign is negative because the loop and current are in the same direction.
01:46I1 R1 minus I2 R2 equals zero.
01:51This equation can be written like this.
01:56We now have two mathematical equations.
02:01From equation 2, I2 is equal to I1 R1 over R2.
02:08Simply substitute the value of I2 from equation 2 into equation 1.
02:16Both factors on the right contain I1, so we can exclude I1 from there.
02:23From this, we get the current I1.
02:29Using the same method, we will get the current I2.
02:35This is the equation for the current in each resistor.
02:40Do you want to memize this formula?
02:43I don't think it's necessary.
02:46We just need to properly understand the concepts of Kirchhoff's current law and Kirchhoff's voltage law.
02:53After this, we just need to check some of the values listed on the problem sheet.
02:59A little calculation, I1 is 8 Amperes.
03:04And I2 is 4 Amperes.
03:10It turns out that the sum of the currents I1 and I2 is 12 Amperes, corresponding to the current entering
03:16the circuit.
03:20Because the current is divided across several resistors, this circuit is often known as a current divider circuit.
03:31Happy learning, everyone!
03:32Thank you for this brings up a congruence I1.
03:34Thank you, everyone!
03:35And if I did this Division I1, please include Q1 on the left.
03:35What is left behind me!
Comments