A circuit contains four light bulbs. One light bulb goes out but the other three stays it. This must be an)
1 answer:
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Answer:
R_total = 14.57 Ω , V_C = 1.176 V
Explanation:
To solve this circuit we are going to find the equivalent resistance of each branch, let's remember
* Serial resistance
= ∑
* For resistance in parallel
1 / R_{eq} = ∑ 1/R_{i}
We solve the two branches of the wheatstone bridge
Series resistors
Branch B
R_B = Rb + R4
R_B = 2 + 18
R_B = 20 Ω
Branch C
R_C5 = Rc + R5
R_C5 = 3 + 12
R_C5 = 15 Ω
Resistance in parallel R_B and R_C5
1 / R_BC = 1 / R_B + 1 / R_C5
1 / R_BC = 1/20 + 1/15 = 0.116666
R_BC = 8.57 Ω
Now we have a single branch, we solve the series resistance
R_total = R_A + R_BC
R_total = 6 + 8.57
R_total = 14.57 Ω
b) they ask us for the voltage in the resistance R_C
Let's remember that the voltage in a series circuit is the sum of the voltages
10 = V_a + V_BC
10 = i R_a + i R_BC = i (R_a + R_BC)
i = 10 / (R_a + R_BC)
i = 10 / (14.57)
i = 0.6863 A
The current in the series circuit is constant
V_BC = i R_BC
V_BC = 0.6863 8.57
V_BC = 5.8819 V
This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.
Branch C
V_BC = i R_C5
i = V_BC / R_C5
i = 5.8819 / 15
i = 0.39213 A
In this branch we have two resistors in series, let's remember that the current in a series circuit is constant
V_C = i R_C
V_C = 0.39213 3
V_C = 1.176 V