Math Problem Statement

A 30-ohm resistor is connected in parallel with a variable resistance R. The parallel combination is then connected in series with a 6-ohm resistor and connected across a 120V source. Find the minimum value of R if the power taken by the 6-ohm resistor.

Solution

To solve this problem, we need to determine the minimum value of RR such that the power consumed by the 6Ω6\Omega resistor is maximized. Here's the step-by-step process:


1. Understand the Circuit Configuration:

  • Parallel combination: 30Ω30 \Omega and RR.
  • Series connection: The parallel combination is connected to a 6Ω6 \Omega resistor and powered by a 120V120V source.

2. Recall Power Maximization Rule:

The power PP dissipated in a resistor is given by: P=V2RP = \frac{V^2}{R} where VV is the voltage across the resistor, and RR is its resistance. To maximize the power in the 6Ω6 \Omega resistor, the voltage across it must also be maximized.

For maximum voltage across 6Ω6 \Omega:

  1. The total current in the circuit should be minimized (reducing voltage drop across the parallel combination).
  2. The equivalent resistance ReqR_\text{eq} of the parallel combination should approach infinity, meaning RR must be very high.

3. Compute Equivalent Resistance of the Parallel Combination:

The equivalent resistance of 30Ω30 \Omega and RR is: Rparallel=30R30+RR_\text{parallel} = \frac{30R}{30 + R}


4. Total Resistance in the Circuit:

The total resistance in the circuit is: Rtotal=Rparallel+6R_\text{total} = R_\text{parallel} + 6


5. Current in the Circuit:

The current II from the source is: I=VsourceRtotal=12030R30+R+6I = \frac{V_\text{source}}{R_\text{total}} = \frac{120}{\frac{30R}{30 + R} + 6}


6. Power in the 6Ω6 \Omega Resistor:

The voltage across the 6Ω6 \Omega resistor is: V6=I×6V_{6} = I \times 6 The power dissipated in the 6Ω6 \Omega resistor is: P6=V626=6I2P_{6} = \frac{V_{6}^2}{6} = 6I^2


7. Maximize P6P_{6}:

For P6P_{6} to be maximized, RparallelR_\text{parallel} must be minimized. The minimum RparallelR_\text{parallel} occurs when RR \to \infty, making Rparallel30ΩR_\text{parallel} \approx 30 \Omega.

With RR \to \infty:

  • Rtotal=Rparallel+6=30+6=36ΩR_\text{total} = R_\text{parallel} + 6 = 30 + 6 = 36 \Omega.
  • Current I=12036=3.33AI = \frac{120}{36} = 3.33 \, \text{A}.
  • Power in 6Ω6 \Omega: P6=6×(3.33)2=6×11.1=66.6W.P_{6} = 6 \times (3.33)^2 = 6 \times 11.1 = 66.6 \, \text{W}.

Conclusion:

The minimum value of RR for maximum power in the 6Ω6 \Omega resistor is infinite. Practically, RR must be very large to approximate this condition.


Questions to Explore:

  1. What happens to the circuit behavior if RR is very small?
  2. How does RparallelR_\text{parallel} affect the overall circuit current?
  3. How would the maximum power change if the supply voltage was reduced?
  4. Can you derive the critical value of RR that gives a specific fraction of maximum power?
  5. What is the efficiency of the circuit under maximum power conditions?

Tip:

To maximize power in any series-parallel combination, focus on balancing resistances to shift the voltage drop toward the resistor of interest.

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Math Problem Analysis

Mathematical Concepts

Electrical Circuit Analysis
Resistors in Parallel and Series
Maximization of Power in Circuits

Formulas

P = V^2 / R (Power formula)
R_parallel = (R1 * R2) / (R1 + R2) (Parallel resistance formula)
I = V / R (Ohm's Law)

Theorems

Maximum Power Transfer Theorem

Suitable Grade Level

Grades 11-12 (Physics and Advanced Math)