Explanation Formula Used: The expression for current by Kirchhoff’s junction rule is given by, I 1 + I 2 = I 3 The expression for the loop rule outside the circuit is given by, ε 1 − I 3 R − r 1 I 1 = 0 …… (I) The expression for the loop rule inside the circuit is given by, ε 2 − I 3 R − r 2 I 2 = 0 I 2 = ε 2 − I 3 R r 2 …… (II) Equation (I) can be re written as, ε 1 − I 3 R − r 1 I 1 = 0 ε 1 − I 3 R − r 1 ( I 3 − I 2 ) = 0 ε 1 − I 3 R − r 1 ( I 3 − ε 2 − I 3 R r 2 ) = 0 I 3 = ε 1 r 2 + ε 2 r 1 r 1 r 2 + R ( r 1 + r 2 ) The expression for the power delivered to R is given by, P = I 3 2 R = ( ε 1 r 2 + ε 2 r 1 r 1 r 2 + R ( r 1 + r 2 ) ) 2 R = ( ε 1 r 2 + ε 2 r 1 r 1 + r 2 ) 2 [ R ( R + A ) 2 ] Here, A = r 1 r 2 r 1 + r 2 The quantity in the parenthesis is independent of R so ignore it

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Author: Mosca

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Chapter 25, Problem 117P

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The proof if a resistor of resistance R is connected in parallel with combination then the optimal load resistance is given R=r1r2/(r1+r2) .

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Chapter 25, Problem 116P

Chapter 25, Problem 118P

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Chapter 27, Problem 039 GO In the figure two batteries of emf E = 11.0 V and internal resistance r = 0.306 are connected in parallel across a resistance R. (a) For what value of R is the dissipation rate in the resistor a maximum? (b) What is that maximum? HF 8 FWW 8 R

A battery with emf E and internal resistance r is first connectedto a resistor of resistance R1 = R, where R is greater than r.This resistor is then disconnected from the battery and replaced with aresistorof resistance R2 = 2R. Find the ratio of the power dissipated in the second resistor to the power dissipated in the first resistor.

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The proof if a resistor of resistance R is connected in parallel with combination then the optimal load resistance is given R = r 1 r 2 / ( r 1 + r 2 ) .

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The proof if a resistor of resistance R is connected in parallel with combination then the optimal load resistance is given R=r1r2/(r1+r2) .

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Chapter 25 Solutions