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;

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

Small charge in the resistance that can be detected.

.

Expert's answer

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

Fredy19.07.21, 10:40Bravo! Thanks so much. To the above question, what is the smallest charge in resistance that can be detected?

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