Question

The circuit to the right consists of a battery (0=1.50 V) and five resistors (1=511 Ω, 2=182 Ω, 3=663 Ω, 4=234 Ω, and 5=565 Ω). Determine the current point passing through each of the specified points.

The battery V subscript 0 is connected in series via its positive terminal with resistor R subscript 1. R sub 1 is connected in series with the parallel combination of resistors R subscript 2, R subscript 3, R subscript 4, and R subscript 5. The parallel resistor combination is connected to the negative terminal of the battery.

The parallel resistor combination is configured as a rectangle with a total of four vertical branches. Each vertical branch contains a resistor. The resistors are in numerical order, with R sub 2 on the leftmost branch and R sub 5 on the rightmost branch. Resistor R sub 1 connects to the parallel combination at the midpoint of the upper horizontal leg, between the central parallel branches containing R sub 3 and R sub 4.

The circuit contains eight points labeled A through H. Point A is located on the upper horizontal leg between the parallel branches containing R sub 2 and R sub 3. Point B is located on the upper horizontal leg between the central parallel branches containing R sub 3 and R sub 4, and is to the left of the junction leading to R sub 1. Point C is located on the same horizontal leg as point B, but is to the right of the junction. Point D is located on the upper horizontal leg between the parallel branches containing R sub 4 and R sub 5. Point H is located on the lower horizontal leg between the parallel branches containing R sub 4 and R sub 5, and is opposite point D. Point G is located on the lower horizontal leg between the central parallel branches containing R sub 3 and R sub 4, opposite the junction leading to R sub 1. Point F is located on the lower horizontal leg between the parallel branches containing R sub 2 and R sub 3, opposite point A. Point E is located between the parallel resistor combination and the negative terminal of the battery.

D=

mA
G=

mA
F=

mA

Answer 1

• D: 0.346 mA
• G: 1.182 mA
• F: 1.478 mA

Step-by-step explanation:

You want the current in various circuit branches of a series-parallel circuit with a battery voltage V0 = 1.5V, a series resistor of R1 = 511 Ω, and four parallel resistors, R2–R5 = 182, 663, 234, and 565 Ω, respectively.

Solution

There are a number of ways to solve the circuit. The one shown in the second attachment finds the combination of the parallel resistors, then determines how the total current is split among them. The values of interest include the current through R5 (node D), the sum of currents through R5 and R4 (node G), and the sum of currents through R5, R4, and R3 (node F).

If Rx is the effective resistance of the parallel combination of R2–R5, then the battery current is

I = V/R = (1.50)/(511 +Rx) ≈ 2.55254 mA

The current in any resistor Rn is this value multiplied by the fraction Rx/Rn for n=2 to 5.

Mesh Currents

Perhaps more directly, we can write "mesh current" equations for the circuit. Letting I1–I4 represent the currents through nodes D, G, F, and E, respectively, we can write the equations ...

• I1(0.565 +0.234) -I2(0.234) = 0
• I1(-0.234) +I2(0.234 +0.663) -I3(0.663) = 0
• I2(-0.663) +I3(0.663 +0.182) -I4(0.182) = 0
• I3(-0.182) +I4(0.511 +0.182) = 1.50

The solution to these equations is shown in the first attachment. Resistances are given in kΩ so currents will be in mA.

The currents in the listed nodes are ...

• D: 0.346 mA
• G: 1.182 mA
• F: 1.478 mA

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Additional comment

The third attachment shows the circuit as we understand it. The currents labeled I1–I4 are within the local loop. The "mesh current" equations match Kirchoff's Voltage Law: the sum of voltage differences around any closed loop is zero. Where a resistor is shared between loops, the voltage across it will be the (signed) sum of the two loop currents times that resistance.