Push of a Cell

Why it matters

A new cell proudly says 6 V. Make it drive a heavy load and the meter across it reads less. Where did the missing volts go?

The idea in plain words

The honest equation of every battery. Tap each term.

V = ε − I·r

ε = 6 V, internal r = 1 Ω, external R = 11 Ω. Find the current and terminal voltage.

V = 5.5 V

• volt (V): V — terminal voltage — what the outside circuit actually gets
• volt (V): ε — e.m.f. — total energy the cell gives each coulomb
• ampere (A): I — current being drawn
• ohm (Ω): r — internal resistance of the cell itself
• I = ε ÷ (R + r) = 6 ÷ 12 = 0.5 A
• V = ε − I·r = 6 − 0.5 × 1

Predict first

A cell of e.m.f. 6 V is pushing a large current through a circuit. Its terminal voltage will read…

Some volts are spent pushing the current through the cell's own insides: V = ε − I·r. The harder the cell works, the more it keeps for itself.

• less than 6 V — correct
• exactly 6 V — a cell always gives its full voltage
• more than 6 V

What you do

Drag the load resistance DOWN to make the cell struggle — watch the red lost volts grow. Then ease off and recover almost all of the e.m.f.

Check yourself

When does the terminal voltage equal the e.m.f.?

With I = 0 there are no lost volts (I·r = 0), so V = ε. That is why e.m.f. is measured on an open circuit.

ε = 12 V, I = 2 A, r = 0.5 Ω. Terminal voltage?

V = 12 − 2 × 0.5 = 11 V.

The volts lost inside the cell equal…

The cell's insides are just another resistor: lost volts = I·r, by Ohm's law.

• When no current is being drawn — correct
• When the current is largest
• Never
• 11 V — correct
• 13 V
• 12 V
• I × r — correct
• ε × r
• I ÷ r