Answer:
0.9 A
Explanation:
You want the current between two voltage dividers in the bridge circuit shown.
Useful relations
When a series circuit of two resistors is supplied by a voltage V, the voltage across one of the resistors will be ...
V₁ = V·R₁/(R₁ +R₂) . . . . . . . where V₁ is the voltage across R₁
This "voltage divider" relation follows directly from Ohm's Law for the series circuit, which tells you the current is V/(R₁ +R₂) and the R₁ voltage is the product of that resistance and the current.
Another useful relation is the equivalent resistance of parallel resistors:
[tex]R_{eq}=R_1||R_2=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}}=\dfrac{R_1R_2}{R_1+R_2}[/tex]
If the resistors are related by R₂ = n·R₁, then this becomes ...
[tex]R_{eq}=\dfrac{R_1\cdot nR_1}{R_1+nR_1}=R_1\dfrac{n}{n+1}[/tex]
Circuit Analysis
Replacing the ammeter (A) by a wire reduces the circuit to a voltage divider with a top resistor of R₂║R₁ = (6·2)/(6+2) = 12/8 = 1.5 ohms, and a bottom resistor of R₁║R₁ = (2·2)/(2+2) = 4/4 = 1 ohm.
Then the voltage divider relation tells you the voltage on across that bottom 1Ω resistance is ...
V = 9(1)/(1+1.5) = 9/2.5 = 3.6 . . . . volts
The current in each resistor can now be found. It will be the ratio of the voltage across the resistor terminals to the resistance: Ohm's Law.
- upper left R₂: (9 -3.6)/6 = 0.9 amperes
- upper right R₁: (9 -3.6)/2 = 2.7 amperes
- lower left R₁: (3.6)/2 = 1.8 amperes
- lower right R₁: (3.6)/2 = 1.8 amperes
Kirchoff's current law (KCL) tells you the current into a node is equal to the current out of it. Considering the node at the right side of the ammeter (A), the current into it is 2.7 A (through upper right R₁). The current out of it through lower right R₁ is 1.8 A, so there must be ...
2.7 A -1.8 A = 0.9 A
out of the middle right node in the direction through the ammeter.
The wire marked A will be carrying 0.9 A.
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Additional comment
That 0.9 A through (A) adds to the 0.9 A through R₂ so that the total into the left-side node between R₁ and R₂ is 0.9 +0.9 = 1.8 A. The current out of that node is the 1.8 A current through the lower left R₁ as we computed above. Hence these values are all consistent.
Each voltage divider can be replaced by its "Thevenin equivalent" circuit. The open-circuit voltage is the voltage at its middle node. On the left, that is 9(2/(2+6)) = 9/4 = 2.25 volts. On the right, that is 9(2/(2+2)) = 9/2 = 4.5 volts.
The equivalent impedance of the left source is the parallel resistance of R₂ and R₁, which we found above to be 1.5Ω. The equivalent on the right is 1Ω. So, the Thevenin equivalent circuit that is seen by (A) has a voltage on the left of 2.25 V with an impedance of 1.5Ω, and a voltage on the right of 4.5 V with an impedance of 1Ω. The current through (A) will be (4.5 -2.25)/(1.5 +1) = (2.25V)/(2.5Ω) = 0.9 A.
Analyzing "bridge" circuits in this way can be easier than other methods (node voltage or loop current).
