Explanation Given: The emf and internal resistance of the two identical batteries is ε and r respectively. Formula used: The expression for the Kirchhoff’s loop rule is given by, ∑ V = 0 The expression for the power supplied is given by: P = I 2 R The expression for the junction rule is given by: I i n c o min g = I o u t g o i n g Calculation: The Kirchhoff’s loop rule is applied to the circuit as, ∑ V = 0 − r I s + ε − r I s + ε − I s R = 0 I s = 2 ε 2 r + R The power supplied to Rin the series circuitis calculated as, P = I 2 R P s = ( 2 ε 2 r + R ) 2 R = 4 ε 2 ( 2 r + R ) 2 The Kirchhoff’s loop rule is applied to the loop 1, ∑ V = 0 − r I 1 + ε + r I 2 − ε = 0 I 1 = I 2 By using the Kirchhoff’s junction rule, I i n c o min g = I o u t g o i n g I p = I 1 + I 2 I p = 2 I 1 Apply the Kirchhoff’s loop rule to the outer loop: ε − R I p − r I 1 = 0 ε − R I p − r ( I p 2 ) = 0 I p = ε ( 1 2 r + R ) The power supplied to Rin the parallel combination is calculated as: P p = ( I p ) 2 R P p = ε 2 ( 1 2 r + R ) 2 R Now suppose R > r ,then 3 R 2 > 3 r 2 Adding r 2 + 4 r R + R 2 on both sides, 3

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Chapter 25, Problem 87P

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Explicitly whether the power supplied to R is greater than the when R is less than or greater than r .

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Chapter 25, Problem 86P

Chapter 25, Problem 88P

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Explicitly whether the power supplied to R is greater than the when R is less than or greater than r .

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Chapter 25 Solutions