Question

The emf of a battery is 2v and its internal resistance is 0.5 ω. the maximum power wch it can deliver to any external circuit will be

Answer 1

applying ohm's law we get I =4
p=v/I
p=0.5

Answer 2

Answer:

The maximum power that can be delivered is 2watt.

Power:

• Power is defined as the transfer of energy per unit of time.
• It can also be called the rate of doing work.
• It is calculated as $P=\frac{W}{dt}$, where W is the work.
• Its SI unit is a watt equal to one joule per second.

Explanation:

Given:

emf (E) = 2V, resistance (R) = 0.5Ω

To find:

Maximum power (P)

Formula:

$P_{max}=\frac{E^{2} }{4R}$

Solution:

Substituting the given values in the above formula, we get

$P_{max}=\frac{2^{2} }{4(0.5)}$

$P_{max}=\frac{4 }{4(0.5)}$

$P_{max}=\frac{1 }{0.5}$

$P_{max}=\frac{10 }{5}$

$P_{max}=2W$

Therefore, the battery can deliver a maximum power of 2watt to any external circuit.

#SPJ2