C/P 04 · Circuits, Ohm's law and membranes
Resistors and Ion Channels
Analyze a series–parallel circuit, then apply the same rules to channels in a membrane.
Passage
Electrical circuits obey two simple rules. Ohm’s law states that the voltage V across a resistor equals the current I through it times its resistance R: V = IR. The power dissipated in a resistor is P = IV = I²R. When resistors are connected in series, the same current flows through each and the resistances add. When resistors are connected in parallel, the voltage across each is the same, and the reciprocals of the resistances add: 1/R = 1/R₁ + 1/R₂.
Meters are used to check such predictions. An ammeter is connected in series with the part of the circuit being measured and a voltmeter is connected in parallel across it. In an ideal circuit, an ammeter has no resistance and a voltmeter draws no current, so neither one changes the quantities being measured. Capacitors store charge according to Q = CV, and when a capacitor is charged or discharged through a resistor, the voltage changes gradually rather than instantly.
Part 1. A 12 V battery of negligible internal resistance is connected to three resistors. Resistor R₁ is in series with the battery. Resistors R₂ and R₃ are connected in parallel with each other, and that pair is in series with R₁. Table 1 shows the resistances.
Table 1. Resistances in the circuit.
| Resistor | Resistance (Ω) |
|---|---|
| R₁ | 2 |
| R₂ | 6 |
| R₃ | 3 |
Part 2. Cell membranes can be treated as electrical components. Ion channels in the membrane act as resistors in parallel, and the lipid bilayer acts as a capacitor with capacitance C. A researcher holds a small patch of membrane at a potential difference of −70 mV and measures the total current when different numbers of identical ion channels are open. The magnitude of the current is recorded in picoamperes (1 pA = 10⁻¹² A). Table 2 shows the results.
Table 2. Total current through a membrane patch held at −70 mV.
| Number of open channels | Current magnitude (pA) |
|---|---|
| 10 | 1 |
| 20 | 2 |
| 40 | 4 |
The patch is held at a fixed voltage by a voltage clamp, an amplifier circuit that injects whatever current is needed to keep the potential from changing. The injected current is therefore equal in size to the current flowing through the open channels. Each open channel passes ions in proportion to the voltage across it, and channels that are closed pass essentially no current. In a real cell, the number of open channels changes constantly as the cell signals, so the electrical properties of the membrane change too.
When the membrane potential is changed suddenly, the voltage across the patch approaches its new value gradually. The time constant of this change is τ = RC, where R is the resistance of the patch. The patch capacitance does not change during the experiment.
Questions
Question 1 of 6
What is the total current drawn from the battery in Part 1?
This would be the current if the total resistance were 12 Ω. The parallel pair has an equivalent resistance of only 2 Ω.
This is the current through R₃ alone. Total current is battery voltage divided by the total resistance of the circuit.
This would be the current if only R₁ were in the circuit (12 V ÷ 2 Ω). The parallel pair adds more resistance in series.
Why the answer is what it is
For R₂ and R₃ in parallel: 1/R = 1/6 + 1/3 = 3/6, so R = 2 Ω. In series with R₁, the total is 2 + 2 = 4 Ω. The current is I = 12 V ÷ 4 Ω = 3 A.
Review: Electrostatics & Circuits · Equations: Charge
Question 2 of 6
What is the current through R₂ (6 Ω)?
This is the current through R₃. It has the same voltage across it as R₂ but half the resistance, so it carries twice the current.
This is the total current, which splits between R₂ and R₃ at the junction.
This would require the full battery voltage across R₂. R₁ takes some of the voltage first.
Why the answer is what it is
The current of 3 A through R₁ produces a drop of 3 × 2 = 6 V, which leaves 6 V across the parallel pair. For R₂: I = 6 V ÷ 6 Ω = 1 A. For R₃: 6 V ÷ 3 Ω = 2 A. The two currents add up to the 3 A total.
Review: Electrostatics & Circuits · Equations: Charge
Question 3 of 6
R₃ is removed from the circuit so that no current flows through that branch. What happens to the total current drawn from the battery?
Removing a branch removes a path for charge. With the battery voltage fixed, a larger total resistance means less current.
The battery voltage is the same, but the total resistance has changed, so the current changes too.
The circuit is still complete through R₁ and R₂, so current still flows.
Why the answer is what it is
Without R₃, only R₁ and R₂ remain in series: 2 + 6 = 8 Ω. The current is 12 ÷ 8 = 1.5 A, which is less than 3 A. Parallel branches provide extra paths, so removing one raises the equivalent resistance.
Review: Electrostatics & Circuits · Equations: Charge
Question 4 of 6
What power is dissipated in R₁ in the original circuit?
This is the voltage across R₁ (6 V), not a power. Power is current times voltage, or I²R.
This is the total power supplied by the battery (12 V × 3 A). R₁ dissipates only part of it.
This would result from using I = 1.5 A, which is the current through neither R₁ nor the battery in this circuit.
Why the answer is what it is
R₁ carries the total current of 3 A, so P = I²R = 3² × 2 = 18 W. This matches V × I = 6 V × 3 A = 18 W.
Review: Electrostatics & Circuits · Equations: Charge
Question 5 of 6
What is the resistance of the patch when 10 channels are open?
This is off by a factor of 1000. Convert 70 mV to 0.07 V and 1 pA to 10⁻¹² A before dividing.
This results from using 70 V instead of 0.07 V. Millivolts must be converted to volts.
This inverts the ratio. Resistance is voltage divided by current, which gives a large number for such a small current.
Why the answer is what it is
R = V/I = (7 × 10⁻² V) ÷ (1 × 10⁻¹² A) = 7 × 10¹⁰ Ω.
Review: Equations: Charge · Membrane Potentials & Equilibrium Gradients
Question 6 of 6
Going from 10 open channels to 40 open channels, the time constant τ = RC of the patch will:
More open channels give more parallel pathways for current, which lowers the resistance, not raises it.
Resistances of parallel channels combine as conductances, which add in proportion to the number of channels. There is no squared dependence.
The capacitance is fixed, but τ also depends on R, and R changes.
Why the answer is what it is
Current at a fixed voltage is proportional to the number of open channels (1, 2, 4 pA), so the resistance is inversely proportional to that number. Going from 10 to 40 channels cuts R to one fourth. With C unchanged, τ also falls by a factor of 4, so the membrane responds to changes in voltage more quickly.
Review: Membrane Potentials & Equilibrium Gradients · Resting Membrane Potential · Equations: Charge