Explore IDNLearn.com's extensive Q&A database and find the answers you need. Get accurate and comprehensive answers to your questions from our community of knowledgeable professionals.
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
#SPJ4
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
