Defibrillator capacitor: charge, capacitance, energy, power
SPH4U Grade 12 Physics · Gravitational, Electric and Magnetic Fields
A trainer-model defibrillator is opened on the electronics bench so the class can meter its charging circuit. Inside, a boost converter feeds the unit's single storage capacitor, and an ammeter spliced into the charging lead shows the converter holding a steady 45.0 mA from the moment the charge button is pressed. The charge tone runs for 8.00 s before the ready light comes on, at which point a high-voltage probe reads 900 V across the plates. Pressing the paddles into the test load then dumps the whole store through the patient circuit, and the oscilloscope capture of the pulse shows the discharge lasting 4.00 ms. Find the charge moved onto the plates, the capacitance that implies, the energy held at 900 V, and the average power delivered during the shock.
Every number in this problem is editable — change any value below and the whole chain recalculates.
- I = 45 mA — Charging current from the boost converter
- t₁ = 8 s — Charging time
- V = 900 V — Voltage across the plates at the end
- t₂ = 4 ms — Duration of the discharge
- (a)the charge moved onto the plates
- (b)the capacitance that implies
- (c)the energy held at 900 V
- (d)the average power delivered during the shock
Current is charge per second, so a steady charging current for a known time is simply charge. Nothing about the capacitor is known yet — this step only counts what went in.
Carried onward at full precision, not this rounded figure.
Capacitance is defined by the pair: the charge it took and the voltage that resulted. C = Q/V is a measurement here, not a lookup on a datasheet.
Carried onward at full precision, not this rounded figure.
Energy needs the capacitance from step 2 and the same 900 V. The ½ is not decoration — the first coulomb went on at almost no voltage and the last one at the full 900 V.
Carried onward at full precision, not this rounded figure.
The same joules, released over milliseconds instead of seconds. This is the entire design idea of the device, and the answer is the reason it is dangerous.
Carried onward at full precision, not this rounded figure.
Therefore the converter moves 0.360 C onto the plates, the capacitor measures 400 μF, it holds 162 J at 900 V — and the 4.00 ms dump delivers that at 40.5 kW, two thousand times the power that charged it.
Why this order
The chain runs in this order because that is the order the quantities become knowable. Charge is the only thing measurable while the capacitor is charging; capacitance falls out of charge and voltage together; energy needs capacitance; power needs energy and a clock. Skip to the end and you are guessing. The step students most often botch is step 3, because two energy expressions look interchangeable — E = ½CV² and E = QV. The second is wrong for a capacitor, and the factor of two it costs is exactly the difference between moving charge across a fixed voltage and moving it onto plates whose voltage climbs as the charge arrives. Adding up V dQ from empty to full is where the ½ comes from.
Then look at what step 4 does to the same 162 J. Held for eight seconds it is 20.25 W, about a dim lamp. Released in four milliseconds it is 40.5 kW — two thousand times the charging power — and that ratio, not the energy, is what stops a fibrillating heart. Energy is what you store; power is what you can do with it, and a capacitor's whole job in engineering is to convert slowly-gathered energy into briefly-enormous power. Camera flashes, spot welders, railguns and the ignition coil in a gasoline engine are the same trick at different scales. It is also why a large capacitor in a switched-off amplifier deserves respect for a good while after the plug comes out of the wall.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.