Balanced and Unbalanced Wheatstone Bridge

BALANCED AND UNBALANCED WHEATSTONE BRIDGE Wheatstone Bridge A Wheatstone bridge connection comprises a specific arrangement of five resistors, where none of the resistors are directly in series or parallel with each other. Typically, the majority of questions focus on balanced Wheatstone bridges. R_{4} l_{2} R_{3} R_{4} I_{1} R w R_{2} Balanced Wheatstone Bridge If there is no current through Rs the bridge is said to be balanced. The bridge is considered to be in a balanced state. KVL between voltage V to x: v - x = i_{1} R_{1} = i_{2}*R_{3} KVL between voltage x to 0: x + 0 = I_{1} R_{2} =1 2 R 4 Dividing equation (1) by (2) (1) ... (2) R_{1}*R_{4} = R_{3}*R_{2} R_{1}/R_{2} = R 3 R 4 Rightarrow R 1 * R 4 Rightarrow R 3 * R 2 R_{1}/R_{2} = R_{3}/R_{4} If there is no current through Rs, the bridge is balanced As no current flows through Rs in balanced WB, circuit can be shown as given figure Equivalent resistance of balanced WB: R eq = (R 1 +R 2 )(R 3 +R 4 ) R 1 +R 2 +R 3 +R 4 www R_{1} www R_{3} R_{2} R_{5} R 1. www R_{5} R_{2} R_{4} R_{2} I_{2} wwww R_{4} Ex. Find the electric current through the segment BC in the circuit shown. Sol This is not a Wheatstone bridge. It would have been a Wheatstone bridge if battery had been connected between A and D. This is a simple circuit with three branches in parallel connection. Equivalent circuit is shown below. Thus, voltage at terminal B: V_{B} = 10V Voltage at terminal C: V_{c} = 0V Current through segment BC: i BC = 10-0 5 =2A I = 2A 60 -ww ww 30 10 V 14 0 70 10V www Ex. Find the equivalent resistance between the points A and D in the circuit shown. Sol. Here R_{1}/n_{2} = R_{3}/n_{4} Thus, this is a balanced Wheatstone bridge. Equivalent resistance of balanced WB R eq = (R 1 +R Z )(R 3 +R 4 ) R 3 +R 2 +R 3 +R 4 dot R eq = 1 pi 5 Omega = ((4 + 3)(6 + 3))/(4 + 6 + 2 + 3) = (6 * 9)/15 = 10/5 * Omega ww www 20 V Ex. Find equivalent resistance between the points A and D in the circuit shown. Sol. This is a different representation of Wheatstone bridge. It can be reframed to usual form as shown in figure R_{1}/R_{2} = R_{3}/R_{4} Thus, this is a balanced Wheatstone bridge. 20 D Equivalent resistance of balanced WB: R eq = (R 1 +R 2 )(R 3 +R 4 ) R 1 +R 2 +R 3 +R 4 R eq = (4+6)*(2+3) 4+6+2+3 R eq = 10 3 Omega = (10 * 5)/15 = 10/3 * Omega Ex. Find equivalent resistance between the points A and D in the circuit shown. Sol. Rearrange the given circuit as follows = Here, R_{1}/R_{2} = R_{3}/R_{4} Thus, this is a balanced Wheatstone bridge Equivalent resistance of balanced WB: R eq =3 Omega R oq = (R 1 +R 2 )(R 3 +R 4 ) R 1 +R 2 +R 5 +R 4 = ((4 + 2)(2 + 4))/(4 + 2 + 2 + 4) = (6 * 6)/12 = 3Omega Ex. Find equivalent resistance between the points A and B in the circuit shown. Sol. Rearrange the given circuit as follows: Thus, circuit between point A and B is a Wheatstone bridge and one more resistance is connected parallel to it. Equivalent resistance of balanced WB: R AB = (R 1 +R 2 )(R 3 +R 4 ) R 1 +R 2 +R 3 +R 4 = ((R + R)(RR))/(R + R + R + R) = R 31 C Equivalent resistance of the circuit: R eq = R* R R+R = R 2 R eq = R 2 Ex. Find current through 50 resistance after interchanging position of 5 2 resistance with the battery in the balanced Wheatstone bridge shown. Sol. After interchanging the position of 5 1 resistance, circuits becomes as shown in the figure The resultant circuit is also a balanced Wheatstone bridge Thus, this is a balanced Wheatstone bridge. R_{1}/R_{2} = R_{3}/R_{4} Here 20 30 20 ww 40 10 V 50 www We know that the current through central resistance for balanced Wheatstone bridge is zero. i = 0 Unbalanced Wheatstone Bridge V-i method to determine the equivalent resistance of a circuit Let's consider the potential at point B to be x and at point C to be y. Current through branch BD: Current through branch CD: Applying KCL at point B, Applying KCL at point C, 3x - v = y ...(1) ... (2) i_{2} = y/2 (x - v)/2 + (x - y)/2 + (x + 0)/2 = 0 (y - v)/1 + (y - x)/2 + (y + 0)/2 = 0 I_{1} = x/2 4y - x = 2v 2y - 2v + y - x + y = 0 20 20 20 Putting value of y from equation (1) in equation (2). 4(3x - v) - x = 2v 12x - 4v - x = 2v 11x = 6v Putting value of x in equation (1), Equivalent resistance of the circuit, (1nv)/11 - v = y x = (6v)/11 R eq = v i = v i 1 *i 2 = v v 2 +yi 2 = (2v)/(x + y) = (2v)/((6v)/11 + (7v)/11) = (22Omega)/43 R eq = 22 13 Omega (7v)/11 = y C Ex. Find the equivalent resistance between A and B for the given circuit diagram. Sol. Assume the potential at point B and C be x and y respectively Current through branch BD: Current through branch CD: Applying KCL at point B, x - v + 2x - 2y + 2x = 0 5x - v = 2y i_{2} = y/z i_{1} = x/1 (x - y)/2 + (x - y)/1 + (x + 0)/4 = 0 y = (5x - v)/2 20 (1) ... (2) Applying KCL at point C, 5y - 2x = 2v Solving equation (1) and (2), Putting value of y in equation (2), 5((5x - v)/2) - 2x = 2v 5(5x - v) - 4x = 4v 25x - 5v - 4x = 4v Equivalent resistance of the circuit, x = (9v)/21 = (3v)/7 R eq = v i 1 +i 2 = v x+yi 2 = V/((3V)/7 + (2V)/7) = 7/5 * Omega y = (5((3V)/7) - v)/2 R eq = 7 5 Omega (y - y)/1 + (y - x)/1 + (y + 0)/2 = 0 21x = 9v = (pi*v)/14 = (4v)/2 2y - 2v + 2y - 2x + y = 0