How Do You Calculate the Variable Resistance for Optimal Battery Charging?

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SUMMARY

The discussion focuses on calculating the variable resistance (R) necessary for optimal battery charging, specifically to achieve 25 Watts absorbed by a 10.5V battery with a 0.035 ohm resistor. The user applied Kirchhoff's Voltage Law (KVL) to derive the equation -13 + 0.02*i + iR + 0.035*i + 10.5 = 0, leading to the current (i) of approximately 2.381 A. The challenge lies in accurately incorporating the 0.035 ohm resistor into the final calculation for R.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Basic knowledge of electrical power equations (P = IV)
  • Familiarity with Ohm's Law (V = IR)
  • Ability to manipulate algebraic equations for circuit analysis
NEXT STEPS
  • Learn how to apply Kirchhoff's Laws in complex circuits
  • Study the relationship between power, voltage, and current in electrical systems
  • Explore techniques for calculating equivalent resistance in series and parallel circuits
  • Investigate the impact of resistance on battery charging efficiency
USEFUL FOR

Electrical engineering students, hobbyists working on battery charging circuits, and professionals involved in power management and circuit design.

jesuslovesu
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Homework Statement


http://img2.freeimagehosting.net/uploads/bf073a5eb1.png
http://img2.freeimagehosting.net/uploads/bf073a5eb1.png

Determine the variable resistance R if the 25 Watts to be absorbed by the battery (0.035 ohm resistor and the 10.5 V element).

Homework Equations


V = IR
P = IV

The Attempt at a Solution



Well I did KVL clockwise current flow
-13 + 0.02*i + iR + 0.035*i + 10.5 = 0
2.5 = 0.02i + 0.035i + iR

Then factoring in 25W=10.5*i
i = 2.381 A which is close to the answer but doesn't factor in 0.035 and I'm not sure how to do it
 
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jesuslovesu said:

The Attempt at a Solution



Well I did KVL clockwise current flow
-13 + 0.02*i + iR + 0.035*i + 10.5 = 0
2.5 = 0.02i + 0.035i + iR

Then factoring in 25W=10.5*i
i = 2.381 A which is close to the answer but doesn't factor in 0.035 and I'm not sure how to do it

The question was to compute R such that you got 25 W dissipated in the 10.5V battery. You computed what the current through this battery must be. Now use that in the equation with i and R you wrote above
 

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