The four arms ABCD of a Wheatstones network have the following resista

1.	The four arms ABCD of a Wheatstones network have the following resistances: AB = 2 Ω, BC=4 Ω, CD= 4 Ω and DA= 8 Ω. A galvanometer of resistance 10 Ω is connected between B and D. Find the current through the galvanometer, when the potential difference between A and C is 5 V.
Answer» The current through the galvanometer is 1.071 A. Explanation: Given that, AB = 2 Ω BC=4Ω CD = 4 Ω DA = 8Ω Galvanometer resistance = 10 Ω Potential difference between A and C = 5 V We need to calculate the potential difference Using KIRCHHOFF's current law at node B $\dfrac{v_{1}-5}{2}+\dfrac{v_{2}-v_{1}}{10}-\dfrac{v_{1}}{4}=0$ $3v_{1}+2v_{2}=50$ ....(I) Using Kirchhoff's law at node D $\dfrac{v_{2}-5}{8}+\dfrac{v_{1}-v_{2}}{10}-\dfrac{v_{2}}{4}=0$ $-9v_{2}+4v_{1}=25$ .....(II) From EQUATION (I) and (II) $35v_{2}=125$ $v_{2}=\dfrac{125}{35}$ $v_{2}=3.57\ V$ Put the VALUE of v₂ in to the equation (I) $3v_{1}+2\times3.57=50$ $v_{1}=\dfrac{50-2\times3.57}{3}$ $v_{1}=14.28\ V$ We need to calculate the current through the galvanometer Using formula of ohm's law $I=\dfrac{V_{2}-v_{1}}{10}$ Put the value into the formula $I=\dfrac{14.28-3.57}{10}$ $I=1.071\ A$ Hence, The current through the galvanometer is 1.071 A. Learn more : Topic : Kirchhoff's law brainly.in/question/8511522