The emf of a battery is 12 volts. When the battery delivers a steady current of 0.5 ampere to a load, the potential difference between the terminals of the battery is 10 volts. What terminal voltage would you expect if the battery were connected to a load with less resistance?

The terminal voltage would increase to be > 10 V

Explanation:

Battery emf = 12 volt

Potential difference at 0.5 ampere load in between the terminals is 10 volts

The formula for terminal voltage is

V = emf - IR

Where:

I = Current = 0.5 A

R = Resistance

Therefore, when V = 10 volt, we have

10 V = 12 V - 0.5 × R₁

∴ R₁ = 4 Ω

When the resistance of the load is lower say R₂ < R₁, then we have

V₂ = 12 V - 0.5 × R₁ = 12 V - 0.5 × R₂ < R₁

= 12 V -<0.5 × R₁

Since V₁ = 12 V - 0.5 × R₁

V₂ > V₁ and the terminal voltage would increase.

V₂ > 10 V.

fghbdt gjdtyg thd tghbtrg htd

step-by-step explanation:

y htf dfh drthf hdt hdthyty hty