Power Developed In A Resistor Given Maximum Power Transfer

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Power Developed In A Resistor Given Maximum Power Transfer

The equivalent circuit of figure 14.1 has:

V_{T} = 35 V, R_{T} = 600 Ω

If the conditions of maximum power transfer exist, determine
**(a)** The value of R_{L}.
**(b)** The power developed in R_{L}.
**(c)** The efficiency of the circuit.

**The strings**:
S_{7}P_{3}A_{32} (force-push).
**The math**:

Pj Problem of Interest is of type *force* (push).

**(a)** Maximum power transfer theorem: to transfer maximum power to a load, the equivalent source resistance and load resistance must be *matched* (equal to each other).

Source voltage = V_{T} = 35 V, source resistance = R_{T} = 600 &ohm, load resistance = R_{L}.

By the maximum power transfer theorem, R_{T} = R_{L} = 600 Ω
**Proof of maximum power transfer theorem**:

Power in load = P_{L} = I_{L}^{2}R_{L}

Where I_{L} = current in load.

I_{L} = V_{T}/(R_{T} + R_{L})

So, P_{L} = [V_{T}^{2}/(R_{T} + R_{L})^{2}]R_{L}

The value of R_{L} that maximizes P_{L} satisfies:

dP_{L}/dR_{L} = 0

dP_{L}/dR_{L} = [V_{T}^{2}(R_{T} + R_{L})^{2} -2V_{T}^{2}R_{L}(R_{T} + R_{L})]/(R_{T} + R_{L})^{4}

So, (R_{T} + R_{L})^{2} - 2R_{L}(R_{T} + R_{L}) = 0

So, R_{L} = R_{T}.
**(b)** V_{L} = [R_{L}/(R_{T} + R_{L})]V_{T} = 35/2 = 17.5 V.

So, P_{L} = V_{L}^{2}/R_{L} = (17.5)^{2}/600 =0.51042 watts = 510.42 mW.
**(c)** Circuit efficiency = (V_{L}/V_{T}) x 100 = (17.5/35) x 100 = 50%.

Math

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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