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Sagot :
The current through the 10 Ω resistor is 0.586A which is connected with a resistor in series combination.
Voltage divides in a Series combination and current divides in a parallel combination.
Let the three resistors joined in parallel be R₁, R₂, R₃
where, R₁= 4 Ω
R₂ = 6 Ω
R₃ = 10 Ω
Given, a battery of 12V and a 2Ω resistor say r is series with the parallel combination.
Equivalent resistance(R) in parallel combination is:
1/R = 1/R₁ + 1/R₂ + 1/R₃
1/R = 1/4 + 1/6 + 1/10
On solving, R = 1.9 Ω
Now, the Equivalent resistance(R) of parallel combination is in series with r = 2Ω
Let Equivalent resistance of Series combination be R'
R' = R + r
R' = 1.9 + 2 Ω
R' = 3.9 Ω
Now let's calculate the voltage drop in the resistor r = 2Ω
v = i × r where, i is the current in r and v is the voltage drop across r
v = 3.07 × 2
v = 6.14V
Voltage drop, V' across the Equivalent resistance(R) in parallel combination = Total voltage - voltage drop in the resistor r
V' = 12 - 6.14 V
V' = 5.86V
Now, the Voltage drop, V' across the Equivalent resistance(R) in parallel combination is same for all the three resistors R₁, R₂, R₃
So, Voltage is same in a parallel combination.
V' = I × R₃
5.86 = I × 10
I = 0.586A
Hence, The current through the 10 Ω resistor is 0.586A
Learn more about Series combination here, https://brainly.com/question/12400458
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