Math Problem Statement

Consider the circuit below, determine the voltages across AC, BC and the current flow through R2.

Solution

To solve the circuit problem, we need to determine the voltages across points AC, BC, and the current flowing through R2R_2. Let's break down the process step by step:

1. Simplifying the Circuit:

  • Resistors R4R_4 and R5R_5 are in parallel. 1Rparallel=1R4+1R5=1100Ω+1300Ω\frac{1}{R_{\text{parallel}}} = \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{100 \, \Omega} + \frac{1}{300 \, \Omega} Solving for RparallelR_{\text{parallel}}.

  • After finding RparallelR_{\text{parallel}}, this resistance is in series with R3R_3, forming a new total resistance between points AA and CC.

    Total resistance between AA and CC will be: RAC=Rparallel+R3R_{\text{AC}} = R_{\text{parallel}} + R_3

  • Resistors R1R_1 and R2R_2 are in series between points BB and CC, so the total resistance between BB and CC is: RBC=R1+R2R_{\text{BC}} = R_1 + R_2

2. Total Circuit Resistance:

The circuit is essentially a combination of two branches (AC and BC) in parallel. The total resistance of the circuit is calculated as: 1Rtotal=1RAC+1RBC\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{AC}}} + \frac{1}{R_{\text{BC}}}

3. Finding the Currents:

Using Ohm's law, the total current in the circuit can be found by: Itotal=VtotalRtotalI_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}} Where Vtotal=20VV_{\text{total}} = 20 \, \text{V}.

4. Current in Each Branch:

Once the total current is found, we can find the current through the ACAC and BCBC branches using the current divider rule.

5. Voltage Across AC and BC:

The voltage across ACAC and BCBC is calculated using Ohm’s law for each branch: VAC=IAC×RACV_{AC} = I_{\text{AC}} \times R_{\text{AC}} VBC=IBC×RBCV_{BC} = I_{\text{BC}} \times R_{\text{BC}}

6. Current Through R2R_2:

The current through R2R_2 is the same as the current in the BCBC branch, since R1R_1 and R2R_2 are in series.

Now let me calculate the values.Here are the calculated values for the given circuit:

  1. The equivalent resistance of R4R_4 and R5R_5 in parallel is 75Ω75 \, \Omega.
  2. The total resistance in the AC branch is 125Ω125 \, \Omega.
  3. The total resistance in the BC branch is 50Ω50 \, \Omega.
  4. The total circuit resistance is 35.71Ω35.71 \, \Omega.
  5. The total current in the circuit is approximately 0.56A0.56 \, \text{A}.
  6. The current in the AC branch is 0.16A0.16 \, \text{A}.
  7. The current in the BC branch (and through R2R_2) is 0.4A0.4 \, \text{A}.
  8. The voltage across AC is 20V20 \, \text{V}.
  9. The voltage across BC is also 20V20 \, \text{V}.

This means that the voltage across both branches is equal, which is consistent since they are connected in parallel.

Do you need more detailed explanations, or do you have any other questions?

Related Questions:

  1. How do you calculate the power dissipated by R2R_2?
  2. What would happen to the total current if R4R_4 was increased?
  3. How would the circuit change if R3R_3 was removed?
  4. What is the total power consumption of the circuit?
  5. How can you verify the current divider rule for this circuit?

Tip:

Always double-check if resistors are in series or parallel before simplifying the circuit to avoid mistakes in calculations.

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Math Problem Analysis

Mathematical Concepts

Ohm's Law
Parallel and Series Circuits
Current Divider Rule

Formulas

V = IR (Ohm's Law)
1 / R_parallel = 1 / R4 + 1 / R5
R_total = R_AC + R_BC
I_total = V_total / R_total
V_AC = I_AC * R_AC
V_BC = I_BC * R_BC

Theorems

Current Divider Rule
Ohm's Law

Suitable Grade Level

Grades 10-12