Question

A 20V battery with an internal resistance of 5Ω is connected to a resistor of x ohms. If an additional 6Ω resistor is connected across the battery, find the value of x so that the external power supplied by the battery remains the same

Answer 1

Answer:

18.46e

Explanation:

E =20V (EMF of the battery)

r = 5Ω ( internal resistance)

Current when x is connected:

i1 = E/(r+x) A = 20/(5+x) A

Current when ( 6+x ) Ω connected to it: ( here 6Ω it connected in series with x)

i2 = E/(r+x+6) = 20/(11+x) A

External power = ( i1 )²x = ( i2)²(x+6)

Now solving this equation,

(20/(5+x))²x = (20/(11+x))²(x+6)

x² + 6x - 25 = 0

It has two roots, 2.83 and -6.83

The negative value is impossible, so the answer is 2.83Ω

In both the cases the external power is 18.46W