8.5.P6

In most of the circuit problems we have done in this class, we have used a battery as a power source. A battery provides a constant voltage difference across its terminals (for as long as it hasn’t used up its stored energy). But in some circumstances (such as powering LEDs or in patch clamping), a more complex device is used that provides a constant current. The basic Kirchhoff principles of circuits still hold so let’s work out what the implications are of having a constant current source (CCS) instead of a battery.

In the figure below are shown three situations: (A) a CCS connected across a single resistor; (B) a CCS connected across a pair of resistors in series; and (C) a CCS connected across a pair of resistors in parallel. The CCS puts out a current of I₀ = 1 μA (= 1.0 x 10^-6 Amps) and all of the resistors have the same resistance: R = 2 kΩ (= 2.0 x 10³ Ohms).

A. Find the voltage drop across each of the resistors in each of the three cases. Explain your reasoning (briefly!). You may either give numerical values (with correct units) or express your answers in terms of I₀ and R.

B. Find the voltage rise across the CCS in each of the three cases. Explain your reasoning (briefly!). You may either give numerical values (with correct units) or express your answers in terms of I₀ and R.

C. Do the series and parallel rules for finding single-resistor-equivalent resistors look the same as when we hooked them up across a battery? That is, does case B look the same to the CCS as a single resistor with resistance 2R and does case C look the same to the CCS as a single resistor with resistance R/2? Explain your reasoning

Joe Redish 10/31/02