Question

a 9 volt battery is connected in a series with resistor The Terminal voltage is found to be 8 volt current through the circuit is measured as 5 ampere what is the internal resistance of the battery ​

Answer 1

Answer:

V=E-Ir

8=9-5r

5r=9-8

r=1/5

r=0.2ohm

Answer 2

ANSWER:-

★ Given:-

• Terminal voltage = 8v
• Current = 5A

★ To Find:-

• Internal Resistance = ?

★ Calculation:-

• E = V - Ir

5r = 9 - 8

5r = 1

r = 1/5

r = 0.2 Ω

Therefore, the internal resistance of the battery = 0.2 Ω.