Question

A battery of 16 V and internal resistance 2 ohm is connected to an external resistance R. Find the value of current so that power in the circuit is maximum​

Answer 1

Answer:

The answer is 2 ohm. This is explained in the attachment.

Answer 2

Given,

• A battery of 16V with an internal resistance of 2Ω.

• An external resistance of R is connected to it.

To find,

Value of R to maximise the power in the circuit.

Solution,

Power is given as V*I.

According to Ohm's Law,

V = IR

Substituting the value of V in the equation for power we have,

Power = $I^{2}R$

Total resistance = (R + 2)Ω

The potential difference is 16V.

Current flowing through the circuit = $\frac{16}{ R + 2}A$

Now, Power = $(\frac{16}{R + 2})^{2}*R$

                     = $(\frac{16 }{R + 2})^{2} *R$

Maximizing the power for the variable external resistance R,

         $\frac{dP}{dR} = 0$

         $\frac{d(\frac{16}{R + 2} )^{2}*R }{dR} = \frac{ ( R + 2 )^{2} - 2( R + 2 )R }{ ( R + 2 )^{4} } = 0$

         $( R + 2 )^{2} - 2( R + 2 )R = 0\\2 - R = 0\\R = 2$

Therefore, for the external resistance, R = 2Ω power in the circuit will be maximum.