Current division rule

Lets start by putting down what we know The voltage across each inductor must be the same (just like the current must be the same to series capacitors). Now the voltage induced in an inductor is $$V = L \frac H\$ then use kirchoff's law \$ i = i_1 + i_2\$ and as you already know \$ i1 = 2 \cdot i_2 \$ you have everything you need to work it out. There is no need to make this complicated. Each inductor with the same voltage applied to it, whether AC or a complicated waveform or not, will have current thru it inversely proportional to its inductance relative to the other inductors. This is just like parallel resistors have relative current thru them inversely proportional to their resistance. The same formula as for parallel resistance can be used to determine inductor current sharing. For example, consider two ideal inductors in parallel, a 10 µH and a 30 µH, both starting with 0 current. The current thru the 10 µH inductor will always be 3x the current thru the 30 µH inductor. You don't need to know anything about the magnitude or waveshape of the voltage to make this statement. You would find the currents of the inductors by simply using current division. I had a similar problem where I had a current source of 25mA and two parallel inductors the first of 60mA and the second of 40mA. the answers using current division were the same as in the back of my book which lead me to believe that you can. I am not entirely sure though, just thought this may help since it worked for ...

• I T = Total Input Current • R 1+ R 2 = Resistors values • I 1 , I 2 = Desired value of currents in the resistors Similarly, the following formulas can be used for AC circuits using CDR. I n = I T x (Z n / (Z 1 + Z 2)) Where Z is the impedances (resistances (like inductive and capacitive reactances “X L& X C” in AC circuits). Example (To Verify the CDR): Find the current passes through both resistors using current divider rule for the given

Contents • 1 Voltage Division • 2 Current Division • 3 Practice Problems • 3.1 Problem 1 • 3.2 Problem 2 Voltage Division When we have a voltage across a string of resistors connected in series, the voltage across the entire string will be divided up among the resistors. We can express the voltage across a single resistor as a ratio of voltages and resistances, without ever knowing the current. In the circuit above, v 1 v = R 1 R 1 + R 2 in the circuit below:

The Current Divider Rule is applicable to parallel circuits, in which the current is divided into the number of parallel branches. The current in each branch may have a different value depending on the specification of branch component. In a parallel circuit, the current has different paths to flow but the voltage across each parallel branch is always equal. So the current divider rule explains how the current is divided among the parallel branch. Table of Contents • • • Current divider circuit Let us consider a 1 and R 2 connected across the supply voltage V s. Let I T be the total current flowing in the circuit, which has two parallel paths via resistor R 1 and R 2. I R1 is the current flowing through R 1 and I R2 is the current flowing through R 2. By At node A, by applying From equation (1), the current I R1 through the parallel branch R 1 is obtained as below, Similarly,from equation (2), the current I R2 through the parallel branch R 2 is obtained as below, Solved Problem 1 Consider two parallel resistors 25Ω and 15Ω are connected across the supply voltage of 40 V. Find the current flowing through each resistor and the total current supplied by the source. In the given problem, R1 = 25 Ω and R2 = 15 Ω, Vs = 40 V. The current flowing through each resistor is obtained as, The total current is obtained as, Solved Problem 2 In the following circuit, calculate the current flowing through each resistor and the total current flowing.

Current Divider In the similar way of the voltage divider, we can calculate the expression for branch current. The voltage drop across in the parallel branches is equal to the source voltage V. Hence, Apply KCL, hence the total current [wp_ad_camp_1] Key Points: From Equation 4 and 5 for calculating current I1, the resistance R2 is in the numerator and for calculating I2, the resistance R1 is in the numerator. Use of current divider Rule: The reason we use current division sometimes instead of Ohm’s law because the current division is simpler and easier. We can simply calculate the current flowing through the resistance. Also, we can use Ohm’s law but Ohm’s law needs the circuit voltage, current, and resistance. Using current divider, we can calculate the current easily.

The current divider rule is used to find the current in the individual circuit when two or more circuit elements are connected in parallel with a current source. When the current source is connected to the circuit elements which are in parallel, the total current flowing in the circuit gets divided in each element according to the resistance or impedance of the circuit element. Thus, the current is divided into all the branches. The voltage across each branch element remains the same because these elements are connected in Current Divider Formula Derivation The current division or current divider rule can be well understood with the following explanation. The current through R 1 and R 2 Resistance can be calculated using Ohm’s law; Current through resistance R 1 Current through resistance R 2 The total equivalent resistance of the parallel circuit is R The total current flowing in the circuit is I ; Putting the value of V in equation (1) we get; Putting the value of V in equation (2) we get; Current Divider Formula Thus, the current in the parallel branch is equal to the ratio of the opposite branch resistance to the total resistance multiplied by the total circuit current. The current divider formula is given below. Illustrative Example on Current Divider Rule : Two resistances 50Ω and 100Ω are connected in parallel with a current source of 50 Current through 50 Ω resistance is I 1 2 Note: The current divider rule is used when two or more circuit elements are connected in...

Our goal is to come up with an expression relating output v o u t v_ R2 start text, R, 2, end text . v o u t = ( v i n 1 R1 + R2 ) R2 v_ v o u t ​ = ( v i n ​ R1 + R2 1 ​ ) R2 v, start subscript, o, u, t, end subscript, equals, left parenthesis, v, start subscript, i, n, end subscript, start fraction, 1, divided by, start text, R, 1, end text, plus, start text, R, 2, end text, end fraction, right parenthesis, start text, R, 2, end text The ratio of resistors is always less than 1 1 1 1 for any values of R1 \text v o u t ​ v, start subscript, o, u, t, end subscript by a fixed ratio determined by the resistor values. This is where the circuit gets its nickname: voltage divider. v o u t = v i n R2 R1 + R2 v_ v o u t ​ = v i n ​ R1 + R2 R2 ​ v, start subscript, o, u, t, end subscript, equals, v, start subscript, i, n, end subscript, start fraction, start text, R, 2, end text, divided by, start text, R, 1, end text, plus, start text, R, 2, end text, end fraction v o u t = 12 V ⋅ 3 k Ω 1 k Ω + 3 k Ω v_ v o u t ​ = 1 2 V ⋅ 1 k Ω + 3 k Ω 3 k Ω ​ v, start subscript, o, u, t, end subscript, equals, 12, start text, V, end text, dot, start fraction, 3, start text, k, end text, \Omega, divided by, 1, start text, k, end text, \Omega, plus, 3, start text, k, end text, \Omega, end fraction v o u t = 12 V ⋅ 3 k Ω 4 k Ω v_ v o u t ​ = 1 2 V ⋅ 4 k Ω 3 k Ω ​ v, start subscript, o, u, t, end subscript, equals, 12, start text, V, end text, dot, start fraction, 3, start text, k, end text, \Omega...