Question

Can I get an answer to this question please

Answer 1

Answer:

  (i) 12 V in series with 18 Ω.

  (ii) 0.4 A; 1.92 W

  (iii) 1,152 J

  (iv) 18Ω — maximum power transfer theorem

Explanation:

(i)

As seen by the load, the equivalent source impedance is ...

  10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

  (36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

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(ii)

The load current is ...

  (12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

  P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

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(iii)

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

  (600 s)(1.92 J/s) = 1,152 J

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(iv)

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

  ((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W