Maximum Power

Electricity and Magnetism Level 3

An electric circuit consists of a battery emf $E=\SI{110}{\volt}$ , and internal resistance is $\SI{5}{\ohm}$ and two resistors connected in parallel to the source as shown in figure above. The value of $R$ corresponding to maximum power across $R$ is $\text{\_\_\_\_\_\_\_\_\_\_}$ .

50\Omega

\frac{50}{11}\Omega

5\Omega

\frac{5}{11}\Omega

2 solutions

Sparsh Sarode
Jun 2, 2016

Taking the resistance R as load resistance, the circuit can be drawn as shown below. The battery is shorted to calculate effective resistance between the terminals across which load resistance is connected.

Effective resistance, $R_{eff}=50||5$

$R_{eff}=\dfrac{50x5}{50+5}=\dfrac{50}{11}$

We know that by maximum power theorem(MPT),

$R_{eff}=R_L=\dfrac{50}{11}$

Bonus Question : what is the value of R so that the power Through the circuit is maximum ?

Sabhrant Sachan - 5 years ago

Since the internal resistance of the battery cannot be changed, the power through the circuit will be maximum when effective resistance $R_{eff}$ (of R and 50ohms) will be equal to the internal resistance of the battery,

$\therefore \dfrac{50R}{50+R}=r=5$

$R=\dfrac{50}{9}$

Sparsh Sarode - 5 years ago

exactly :) !

Sabhrant Sachan - 5 years ago

@Sabhrant Sachan – Thanks.. :)

Sparsh Sarode - 5 years ago

Rishi K
Jun 2, 2016

In the diagram,current in resistance R =( E/(r+(50R/50+R))) 50/(50+R) its said that power across R is maximum P=I^2 R, substituting the value of current n equating dp/dr=0 (max power) we get value of R as 50/11.

$I=\Bigg(\dfrac{E}{r+\frac{50R}{50+R}}\Bigg) \Bigg(\dfrac{50}{50+R}\Bigg)$

Replace your formula with this to be more clear... :)

Sparsh Sarode - 5 years ago

Maximum Power

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