In most of the circuit problems we have done in this class, we have used a batte

Question

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 hasnt used up its stored energy). But in some circumstances (such as powering LED5 or in some neuroscience examples) a more complex device is used that provides a constant current. The basic Kirchhoff principles of circuits still hold so lets 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 Io = 1 MuA (= 1.0 x 10^-6 Amps) and all of the resistors have the same resistance: R = 2 kOhm (= 2.0 x 10^3 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 Io and R. B. Find the voltage drop 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 Io 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.

Explanation / Answer

a) By kirchoffs law, along a closed circuit when we move along the direction of current its value is taken is negative thus

V - Ia*Ra=0

=1*10-6*2*103

=2*10-3V ans

b)in resistors in series same current flows and voltage across each resistor differ.thus

VB1= VB2=IB1*RB1 = 1 * 10-6 * 2 * 103 = 2 * 103 V ans

In pallel arrangements

In parallel arrangement potential difference across each resistor is same thus

Vc1=Vc2 = Ic1* Rc1

= 1 * 10-6 * 2*103

= 2 * 10-3 V

In parallel arrangement potential difference across each resistor is same thus

Vc1=Vc2 = Ic1* Rc1

= 1 * 10-6 * 2*103

= 2 * 10-3 V

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