Equivalent Resistance

EQUIVALENT RESISTANCE AND CURRENT DISTRIBUTION IN CIRCUIT Equivalent resistance MRMRMR R_{1} R_{3} \mathfrak{t}_{2} n_{2} R_{3} I_{2} i_{0} Applying Kirchhoff's junction law at point P In general, V R eq = V R 1 + V R 2 + V R 3 1 R eq = 1 R 1 + 1 R 2 + 1 R 3 i_{0} = i_{1} + i_{2} + i_{3} 1 R eu = 1 R 1 + 1 R 2 + 1 R 3 +***...........+ 1 R n ww ww Conclusions: (about parallel combination) 1. Potential difference across each resistor is same. V 2. 1= l_{1} + l_{2} + l_{3} +***........l n . 3. Effective resistance (R) then 1 R = 1/R_{1} + 1/R_{2} + 1/R_{5} +***.........+ 1 R n . (R is less than each resistor). 4. Current in different resistors is inversely proportional to the resistance. 1, etc. where G = 1/R = Conductance of a resistor. [Its unit is Omega ^ - 1 or U(mho) * I I 1 = G 1 G_{1} + G_{2} +***,,...+G n 1. B Ex. In the circuit shown below, find the equivalent resistance between A and B. 60 30 40 40 20 B Sol. Consider the loop enclosed in blue box first. Resistances 4 Ω and 2 2 are connected in series and 3 2 is connected parallel to it. Thus, equivalent resistance of this loop is, Now, replacing this loop with a 20 resistance new circuit will look the adjacent figure. Resistances 40 and 20 are A connected in series and 6 A is connected parallel to it. Thus, equivalent resistance of this loop is, R eq = 6*6 6+6 =3 Omega R eq =3 Omega (6 * 3)/(6 + 3) = (6 * 3)/9 = 2Omega Ex. In the circuit shown below, find the equivalent resistance between A and B. Sol. If we observe carefully, we can find that there are four branches, each having two resistances in series between point A and B. After adding the resistances in series in the branches, given circuit can be represented as shown in figure. The equivalent resistance of the simplified circuit is R eq = 10 3+2 =2 Omega R eq =2 Omega 1 R eq Rightarrow 1 10 + 1 10 + 1 5 + 1 10 Rightarrow 1 R eq Rightarrow 3 10 + 1 5 20 40 The process to determine the equivalent resistance in a complex circuit involves the following steps. 1. Suppose that there is a potential difference V across the terminals of the provided circuit, with one terminal at a potential of 0. 2. Consider two other unknown potentials, denoted as x and y, within the given circuit. R R R 3. Designate V and 0 as the endpoints, and place x between them, on the given scale. R 4. Position all the resistances between the designated points and proceed to solve the circuit. Ex. In the circuit shown below, find the equivalent resistance between A and B. Sol. Here, three resistances R are connected in parallel through circular loop between point x to point B. Resistance R is connected in series with above combination between point A to x. The circuit can be represented as shown in the figure. Thus, equivalent resistance is, R eq =R+ 1 1 R + 1 R + 1 R =R+ R 3 = 4R 3 R eq = 4R 3 R Ex. In the circuit shown below, find the equivalent resistance between A and B. 20 Sol In an 'ideal' short circuit, there is no resistance and thus no voltage drop across the connection. The whole current will flow through the wire having no resistance. Here, one of the three branches between point A to point B is short circuited. Thus, all current will flow through this branch and equivalent resistance of the circuit will be zero. This can also be calculated from formula as follows. R eq =0 Omega 1 n AB = 1 2 + 1 4 + 1 0 Rightarrow R AB =0 A R = 0 B Ex. n the circuit shown below, find the equivalent resistance between A and B. Sol. Here, one of the three branches is short circuited in the part of circuit enclosed in yellow circle. Thus, all current will flow short circuited conductor and equivalent resistance of the part will be zero. The equivalent circuit can be represented as shown in figure. Equivalent resistance of the circuit is, = 2 + 6/3 = 4Omega R eq =2+ 1 1 3 + 1 6 R eq =4 Omega 20 60 B 20 30 60 A Ex. In the circuit shown below, find the equivalent resistance between A and B. Sol Number the resistances as shown in figure. Resistance 1 and 2 are connected in parallel combination between point A to x. Similarly, resistance 3 and 4 are connected in parallel combination between point x to B. The circuit can be represented as shown in the figure. Equivalent resistance of the circuit is, 1 R eq = 1 R + 1 R* R R+R + R* R R+R = 1 R + 1 R 2 + R 2 = 1 R + 1 R R eq = kappa 2 R eq = R 2 Ex. In the circuit shown below, find the equivalent resistance between A and B. 120 40 V30 60 Sol. Number the resistances as shown in figure. From figure, all resistances are connected in parallel. The circuit can be represented as shown in the figure. Equivalent resistance of the circuit is, R eq =1 Omega 1 R eq = 1 3 + 1 6 + 1 12 + 1 4 + 1 6 = (4 + 2 + 1 + 3 + 2)/12 = 1Omega Ex. In the circuit shown below, find the equivalent resistance between A and B. Sol. The circuit can be represented as shown in the figure. Equivalent resistance of the circuit is R eq =6 Omega R eq = a*0 a+0 + 3*6 3+6 = 4 + 2 = 6Omega Ex. In the electrical circuit shown below, find the current in 20 resistance. 30 20V 20 ZA JA 40 20 V 2017 200 30 av Sol. Method 1: Using current distribution Equivalent resistance of the circuit, R AB =5 Omega Current through battery, I' = 20/5 = 4A The 4A current divides into two branches. The 2A current further divides in two branches. Let's assume these currents be in and iz as shown in figure. Ratio of currents, Current through 20 resistance, 11:2 Method 2: Using potential difference across resistor Equivalent resistance of the circuit, R AB =S Omega Current through battery, i' = 20/5 = 4A The voltage drop across the resistance is found by multiplying current with resistance. The voltage at all points is shown in the figure. The voltage drops from 4V to 0 through 40 and 20 resistance. Thus, current through 20 resistance, i = (4 + 0)/(4 + 2) = 4/6 = 2/3 * A i_{1} = 1/3 * 2 = 2/3 i = epsilon/3 * A i = 2/3 * A 20 VT 20 4V 2A 4/6 = (2A)/3 Ex. In the electrical circuit shown below, find the current in 20 resistance. Sol. Equivalent resistance of the circuit, R AB =5 Omega Current through battery. I' = 20/5 = 4A Current through 30 resistance, i_{3\Omega} = (a + 0)/3 = 6/3 * A Let the current through 20 resistance l ^ * . Applying Kirchhoff's current law, i^ prime prime = 4 - 0/3 = 4/3 1^ prime prime + pi/3 = 4 i = 4/3 * A 3日 40 20V Ex. In the electrical circuit shown below, find the current in 30 resistance Sol. Equivalent resistance of the circuit, R eq =4 Omega Current through battery. i_{m} = 20/4 = 5A Let the current through branches from A to B be i_{2} and i 3^ * i_{1} Using short trick, we can write, Therefore, i_{1}; i_{2}; i_{3} = 1/6; 1/3; 1/2 i_{2} = y/(x + y + z) * (i_{m}) = 2/(1 + 2 + 3) * 5 = 5/3 * A = (1/2) / 3 i = 5/3 * A 20 V 60 30 20 20 του Ex. In the electrical circuit shown below, find the value of i. 60 UE 12V Sol. Equivalent resistance of the circuit, R eq =4 Omega Current through battery, Simplified circuit: 12V 20 i_{m} = 3A I_{m} = 14/4 = 3A Here point x represents point C and B on original diagram. Potential at point C and B: Current in branch AB: Current in branch BO: V_{B} = V_{C} = 12 - 3 * 2 = 6V I AB = 12-6 4 =1.5 A i BO = 6-0 6 =1A Applying KCL at point B: 1.5 = i + 1 = i = 0.5A Ex. Find current passing through the battery and each resistor. Sol. Method (1) It is easy to see that potential difference across each resistor is 30 V. 30V + Current is each resistors are 30/2 = 15A, 30/3 = 10A And 30/6 = 5A Current through battery is = 15 + 10 + 5 = 30A . Method (II) By ohm's law Now distribute this current in the resistors in their inverse ratio. i= v R eq Rightarrow 1 R eq = 1 2 + 1 3 + 1 6 =1 Omega R eq =1 Omega Rightarrow i= 30 1 =30A 2 ww 15A 2 A BA 30A 15A nằm Current total in 32 and 6 is 15 A it will be divided as 10 A and 5 A. The method (1) is better. But you will not find such an easy case everywhere Ex. Find current which is passing through battery. Sol. Here potential difference across each resistor is not 30 V Battery has internal resistance. Here the concept of combination of resistors is useful. R eq =1+1=2 Omega i = 30/2 = 15A Ex. Find equivalent Resistance R R R B B 30V, 12