Answer to Question #218291 in Electrical Engineering for Fredy

Question #218291

Circuit diagram of a wheatstone brigde in which Resistance 1=Resistance 2=Resistance 3=Varrying Resistance 4 = 1000ohms and Emf =20volts. The galvanometer can detect as low as 0.1milliamperes. Determine;

  1. Internal resistance of the bridge being measured across the terminal.
  2. Output voltage due to unbalanced conditions.
  3. Current through the galvanometer for unbalanced conditions.

Small charge in the resistance that can be detected.

.




1
Expert's answer
2021-07-19T03:15:53-0400

1.

"V= I(R_a+R_x)\\\\\n\\frac{V}{I}=R_a+R_x\\\\\n\\frac{20}{0.1*10^{-3}}=1000+R_x\\\\\nR_x=199000 \\Omega"

2.

"v= \\frac{R_1}{R_3}*V_1 = \\frac{R_2}{R_4}*V_1= \\frac{1000}{1000}*20=20V"

3.

"I= \\frac{V}{R}\\\\\nI= \\frac{20}{1000}\\\\\nI=0.02A"


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Comments

Fredy
19.07.21, 10:40

Bravo! Thanks so much. To the above question, what is the smallest charge in resistance that can be detected?

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