Calculators/Physics/Ohm's Law for a Circuit (EMF)

Ohm's Law for a Circuit (EMF)

I = ε/(R+r) · terminal voltage · internal resistance

Ohm's Law for a Circuit (EMF) — illustration
Current I
1
A
Terminal voltage U
4
V
Drop across r
0.5
V
Power P = I²R
4
W

Ohm’s law for the circuit: I = ε/(R+r). The terminal voltage U = ε − I·r is smaller than the EMF — part of the voltage is "lost" on the cell’s internal resistance (the more so, the larger the current).

A real current source has not only an EMF (ε\varepsilon) but also an internal resistance (rr). Ohm's law for the whole closed circuit is:

I=εR+rI=\frac{\varepsilon}{R+r}

where RR is the external resistance (the load). The terminal voltage of the source is smaller than the EMF by the drop across the internal resistance:

U=I R=ε−I rU=I\,R=\varepsilon-I\,r

The larger the current drawn, the more voltage is "lost" inside the source. At a short circuit (R=0R=0) the terminal voltage falls to zero and the current reaches its maximum ε/r\varepsilon/r. This is why an old battery (large rr) powers a load poorly.

Examples

ε = 4.5 V, r = 0.5 Ω, R = 4 Ω
→ I = 4.5/4.5 = 1 A, U = 4 V (drop across r = 0.5 V)
Short circuit (R = 0)
→ I = ε/r (short-circuit current), U = 0

Frequently asked questions

How does EMF differ from terminal voltage?+

The EMF is the source voltage with no load. When a current flows, part of the voltage is "lost" on the internal resistance (I·r), so U = ε − I·r remains at the terminals, less than the EMF.

Why does an old battery work poorly?+

Because its internal resistance r grows over time. With a larger r the I·r drop is bigger, so the terminal voltage sags heavily under load even though the EMF stays similar.

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