3. A series circuit offers one (1) pathway for the current.
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Compare the series circuit attached with the circuit you show for problem 4. The attached diagram shows the same current going through each element. In the circuit of problem 4, there are two paths for the current: some current goes through R1 and R2. Other current goes through R1 and R3.
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4. Consider the description above of current flow in this circuit. Call the current through R1 and R2 "i2". Call the current through R1 and R3 "i3". Then the total current through R1 is i2+i3.
The voltage drop requirements are ...
- (i2+i3)R1 + i2·R2 = 10
- (i2+i3)R1 + i3·R3 = 10
Filling in the values of the resistors, these equations become ...
- i2·(2+4) + i3·2 = 10
- i2·2 + i3·(2+10) = 10
Solving these simultaneous equations by your favorite method gives ...
(i2, i3) = (25/17, 10/17) ≈ (1.47, 0.59) . . . amperes
The current through R3 is i3, so is about 0.59 A.
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Alternate solution
The parallel combination of R2 and R3 has the value ...
Rp = (R2·R3)/(R2+R3) = 40/14 Ω = 2 6/7 Ω
Then the total circuit resistance is ...
R1 + Rp = 2 + 2 6/7 = 4 6/7 . . . ohms
and the total circuit current is ...
(10 V)/(4 6/7 Ω) = 35/17 A
When resistors are in parallel, the current divides through them in proportion to the inverse of the resistance value. That is the larger resistor will get the smaller share of current. Its proportion will be 4/(10+4) = 2/7 of the total circuit current of 35/17 A.
R3 current = (2/7)·(35/17 A) = 10/17 A ≈ 0.59 A
