A resistor R1 is wired to a battery, then resistor R2 is added in series. (a) Is the potential difference across R1 now m

Question

A resistor R1 is wired to a battery, then resistor R2 is added in series.

(a) Is the potential difference across R1 now more than, less than, or the same as previously?

(b) Is the current i1 through R1 now more than, less than, or the same as previously?

(c) Is the equivalent resistance R12 of R1 and R2 more than, less than, or the same as R1?

A. more than

B. less than

C. same as

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Vodka 5 months 2021-08-18T16:34:22+00:00 1 Answers 28 views 0

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    2021-08-18T16:36:12+00:00

    Answer:

    a) The voltage on resistor R1 is less than before. Letter B.

    b) The current through R1 is less than before. Letter B.

    c) The resistance of the circuit went up by adding R2 in series. Letter A.

    Explanation:

    a) The new resistor R2 was added in series, that means the same current goes through both resistors (R1 and R2) and that the sum of the individual voltage drop of those resistors should be the same as the voltage of the source. Before there was only resistor R1 connected to the source, so all the voltage of the source was being delivered to that component, but now it is divided between R1 and R2. So the voltage on R1 is less than previously. Letter B.

    b) A battery works as a voltage source, so assuming it’s charged, it delivers a value of voltage to the circuit, while the current flow through it is defined by the equivalent resistance across the terminals. Since the new resistor was added in series the total resistance of the circuit went up and the current delivered by the battery went down. Letter B.

    c) Since the equivalent resistance for a series connection is the sum of the resistors, R12 = R1 + R2, the resistence went up. Letter A.

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