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Circuitry Problems

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1) In the following circuit, all bulbs are identical. The voltage drop across bulb A is 10.0 volts

A) Rank the bulbs in terms of brightness, starting with the brightest.
B) Find the voltage between points 1 and 2. State: "the voltage at point 1 is ________ volts relative to point 2." (In other words, if I use my voltmeter, connecting the black lead at point 2 and the red lead at point 1, what would the voltage reading be?)

2) In the circuit below, the current through the 15-ohm resister is .20 A. Find the voltage drop across the 40-ohm resistor. B) Find the voltage V. C) Find the total current supplied by the battery.

The circuit shown is sometimes called a "bridge" configuration. The current through two of the resistors is indicated. We will call the negative battery terminal our "zero" voltage reference, or "ground".

A) Find the battery voltage. (Hint: what are the voltages drops across the two right-handed resistors?).
B) Find the current through the 3.73-ohm resistor.
C) Find the potential, relative to ground, at point B.
D) Find the potential, relative to ground, at point A. (Hint: you will want to use the voltage at point B to solve this.)
E) Find the current through the 4.0-ohm resistor.
F) Find the total current supplied by the battery.
G) Suppose we replaced the bridge circuit with a single resistor, and that the current through the battery remained the same. What would the value of this "equivalent resistor" be?

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9.7
(A) This we can answer by considering the current through each bulb. The brightest bulb is A because whatever currents are flowing through the B-C string and the D-E-G string, they both must flow through bulb A. Now the B-C string and the D-E-G string are in parallel, so each has the same voltage. But the D-E-G string has more resistance that the B-C string because it has three bulbs instead of two. So less current flows in the D-E-G string than in the B-C string. Thus the bulbs in the B-C string are brighter than those in the D-E-G string. So the ranking is: A, B=C, D=E=G.

(B) The problem states that the voltage drop across bulb A is 10 Volts, and that all bulbs are identical. This tells us that there are 6 Volts across the B-C and D-E-G strings (16 - 10 = 6). Thus there are 3 Volts across bulb B. So if we call the minus terminal of the battery "ground", then with respect to ground the voltage at ...

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