show work. Match the term to its correct description/expression: 1. G Ideal Volt
ID: 2249619 • Letter: S
Question
show work.Match the term to its correct description/expression: 1. G Ideal Voltage SourceP Provides a method for simplifying a circuit to a standared current source equivalent form. EX Thevenin's Theorem 0 X B nimco A X E Norton's Theorem D. Used to precisely mesure res . Has an infinitely large intermal parallel resistance. Provides a method to determine currents in a circuit with multiple sources. y measr resisdinge SAMPLE t Ideal Current Source E. For any given load resistance to two sources, the same load voltage and load current are produced by both sources. Provides a method for simplifying a circuit to a standard voltage source equivalent form. Has a zero internal resistance. E 0Terminal Equivalency 7. C Superposition Theorem 8) / What value does R2 have to be to make the wheatstone Bridge circuit balanced in Figure 1 below. R3 = 1.6 R1 = A B R4 = 7.2 k R2 = Figure 1 Given a source voltage of 66 Vic in Figure 1 above and balanced' with R2 from #8 above, calculate the voltage dropped across R2.
Explanation / Answer
Answer:-1) Matches to G, Ideal voltage source always has zero internal resistance.
2) Matches to F, Thevenin's theorem reduces the total circuit into a single voltage source(called Thevenin's Voltage), series with a resistance(called Thevenin's Resistance) with load resistance.
3) Matches to D, wheatstone bridge is used to measure the resistance value of an unknown resistor.
4) Matches to A, Norton's Theorem reduces the whole circuit into one with a single current source with a parallel Norton resistance.
5) Matches to B, Ideal current source always has the infinite internal parallel resistance.
6) Matches to E, If a resistance is connected to two sources, both sources produces the same load current and voltages.
7) Matches to C, if a circuit has more than one source then the sum of the effect of each sources taken one at a time gives the equivalent value.
8) When a wheatstone bridge is balanced then,
R1/R2 = R3/R4 , keeping the values we get R2 = 13.5 M-ohm.
9) R1 and R2 is in series, so equivalent is R12 = 13.5 + 3 M-ohm = 16.5 M-ohm.
R3 and R4 are in series, so equivalent is R34 = 7.2 + 1.6 k-ohm = 8.8 k-ohm.
Now both equivalents are in parallel, so both has same potential, thus current through R12 is = 66/16.5 micro-A = 4 micro-A. Hence voltage drop across resistance R2 = 4 x 10-6 x 13.5 x 106 V = 54 V.
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