Question

Help with numbers 3 and 4

Answer 1

Three

Series circuits allow only one way to travel. You start at one end of the resistors in series and you wind up at the other end. There is no second path.

Answer:B

Four

Find the parallel resistance first.

Rt(2,3) = (R2 * R3) / (R2 + R3)

Rt(2,3) = 4*10/(4 + 10) = 40/14 = 2.857

Find the total resistance of the circuit

Rt = Rt(2,3) + R1

Rt = 2.857 + 2

Rt = 4.857

Find the current of the entire circuit.

I = E / Rt

I = 10 / 4.857

I = 2.059

Find the voltage drop across R1

R = 2

I = 2.059

E = I * R

E = 2.059 * 2

E = 4.118

Find the voltage across the 2 parallel resistors.

Ep= 10 - 4.118

Ep =5.882 volts across the parallel circuit

Find the current going through R3

R3 = 10 ohms

Ep = 5.882

I3 = E/R

I3 = 5.882/10

I3 = 0.5882 amps

Answer 2

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