What potential difference is measured across a 19.9 Ω load resistor when it is connected across a battery of emf 2.46 V and internal resistance 0.561 Ω? Answer in units of V.

The potential difference measured is 2.388 V.

Explanation:

E = I (R + r)

E is the emf (electromotive force) of the battery = 2.46 V

R is the resistance of the resistor = 19.9 ohms

r is the internal resistance = 0.561 ohm

I (current) = E / (R + r) = 2.46 / (19.9 + 0.561) = 2.46/20.461 = 0.12 A

V (potential difference) = IR = 0.12 * 19.9 = 2.388 V

2.388 V

Explanation:

Using

E = I (R+r) ... Equation 1

Where E = emf, I = current, R = External resistance, r = internal resistance

I = E / (R+r) ... Equation 2

Given: E = 2.46 V, R = 19.9 Ω, r = 0.561 Ω substitute into equation 2 to get the current

I = 2.46 (19.9+0.561)

I = 2.46/20.461

I = 0.12 A.

From Ohm's Law,

V = IR ... Equation 3

Where V = Potential difference across the 19.9 Ω resistor

Substitute the value of I and R into equation 3

V = 19.9 (0.12)

V = 2.388 V