Questions: OB Gremit challenge SOLVING FOR DC CIRCUIT COMPONENT VALUES

OB Gremit challenge SOLVING FOR DC CIRCUIT COMPONENT VALUES
Transcript text: OB Gremit challenge SOLVING FOR DC CIRCUIT COMPONENT VALUES
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Solution

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Solution Steps

Step 1: Identify the circuit configuration

The circuit consists of a current source, four resistors (R1, R2, R3, and R4), and a voltage source. Resistors R2, R3, and R4 are connected in parallel, and this parallel combination is in series with R1. The voltage source is connected in parallel with the entire resistor network.

Step 2: Calculate the equivalent resistance of the parallel resistors

Let Rp be the equivalent resistance of the parallel combination of R2, R3, and R4. The values are R2 = 190 Ω, R3 = 204 Ω, and R4 = 193 Ω. The formula for equivalent resistance of parallel resistors is:

1/Rp = 1/R2 + 1/R3 + 1/R4

1/Rp = 1/190 + 1/204 + 1/193 1/Rp ≈ 0.00526 + 0.00490 + 0.00518 1/Rp ≈ 0.01534 Rp ≈ 1/0.01534 Rp ≈ 65.17 Ω

Step 3: Calculate the total resistance of the circuit

The equivalent parallel resistance (Rp) is in series with R1. R1 = 248 Ω. The total resistance (Rt) is the sum of these two resistances:

Rt = R1 + Rp Rt = 248 + 65.17 Rt ≈ 313.17 Ω

Step 4: Calculate the current through the circuit

The voltage source has a value of 17.12 V. Using Ohm's law (V = IR), we can calculate the current (I) flowing through the entire circuit:

I = V / Rt I = 17.12 / 313.17 I ≈ 0.0547 A or 54.7 mA

Final Answer

The current flowing from the current source is approximately \\(\boxed{54.7 \text{ mA}}\\). This is because the current source and the voltage source are in parallel with the entire resistor network. Therefore the ammeter connected to the current source reads approximately 54.7 mA.

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