A wheatstone bridge ABCD is arranged as follows AB=1 ohm ;BC= 2ohm CD=3ohm; DA=4ohm; . A resistance of 5ohm is connected between B and D. A 4volt battery of internal resistance 1ohm is connected between A and C .Calculate 1. the magnitude and direction of current in 5ohm resistor and 2. the resistance between A and C
Answers
Answer:
NOW, cramer's rule say that: l1=lx/l, l2=ly/l, l3=lz/l. That is....
l1= 550÷230=55/23A
l2= 200÷230=20/23A
l3= 50÷230=5/23A
Voltage across BD,
V=l3×5=5/23×5=10/23= 0.435 ..... Answer...
give me brainlist answer plz......
Answer:
(i) The current in the 5 ohm resistor is 0.087A and the direction is from B to D.
(ii) The resistance between A and C is .
Explanation:
Given a Wheatstone bridge ABCD.
The Wheatstone Bridge is a simple circuit used for measuring an unknown resistance by comparing its value with that of known resistance.
Applying KVL in loop ABDA
...........(1)
Applying KVL in loop BCDB
........(2)
Applying KVL in FABCEF
...........(3)
Equation (1)×2-(2)
⇒ ...........(4)
Equation (1) X 4-(3)
........(5)
Equation (5)×4-(4)×17
⇒
(i) The current in the 5 ohm resistor is 0.087A and the direction is from B to D.
, substituting in equation (4) gives
Substituting I_3 and I_2 in equation (2) we get
Current supplied by battery,
Potential difference or voltage between A and C = EMF of battery - Drop in battery.
Resistance between A and C = Voltage across AC Battery current
(ii) Therefore, the resistance between A and C is .
To know more about Wheatstone bridge go to the following link.
https://brainly.in/question/84419
To know more about current go to the following link.
https://brainly.in/question/1151879
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