Question

Q10. A 120V has a series internal resistance of 1
ohm. The maximum power that can be
delivered to a load is
(A) 1.900 W
(B) 2.1800 W
(C) 3.2700 W
(D) 3600 W​

Answer 1

Answer:

The correct answer is (D) $3600$W​.

Step-by-step explanation:

Given: A $120V$ has a series internal resistance of $1$.

ohm

To find: the maximum power.

Concept:

According to the maximum power theorem for DC circuits, maximum power is delivered to the load when the load resistance $\left(R_{L}\right)$ equals the Thevenin equivalent resistance across $(Rth)$ i.e.

The maximum amount of power is delivered to the load when $\mathbf{R}_{\mathbf{L}}=\mathrm{R}_{\mathrm{th}}$

The maximum power transferred is given by

$P_{\max }=\frac{V_{t h}^{2}}{4 R_{t h}}$

Calculation:

Given:

$V_{\text {th }}=120 \mathrm{~V}$

$R_{\text {th }}=1 \Omega$

The Thevenin equivalent circuit can be drawn attached file:

It is now possible to compute the maximum power delivered to the load as

$&P_{\max }=\frac{120^{2}}{4 \times 1}$

$&P_{\max }=3600 \mathrm{~W}$

Derivation: Attached file.

The current through the load in the circuit above, denoted by the letter I is supplied.as

$I=\frac{V_{t h}}{R_{L}+R_{T H}}$

Power transferred to the load:

$P_{L}=I^{2} R_{L}=\left(\frac{V_{\text {th }}}{R_{L}+R_{T H}}\right)^{2} \cdot R_{L}$

The condition for maximum power given to the load is established by differentiating load power with regard to load resistance and equating it to zero in the equation above, where $\mathrm{R}_{\mathrm{L}}$ is a variable.

$\frac{\partial P_{L}}{\partial R_{L}}=V_{T H}^{2}\left[\frac{\left(R_{L}+R_{T H}\right)^{2}-2 R_{i}\left(R_{T H}+R_{L}\right.}{\left(R_{L}+R_{T H}\right)^{4}}\right]=$

$&\left(R_{L}+R_{T H}\right)^{2}=2 R_{L}\left(R_{T H}+R_{L}\right)$

$&R_{L}+R_{T H}=2 R_{L}$

$&R_{L}=R_{T H}$

The required answer is (D) $3600 W$.

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