PLEASE EXPLAIN ALL ANSWERS Consider the circuit at right, consisting of two iden

Question

PLEASE EXPLAIN ALL ANSWERS Consider the circuit at right, consisting of two identical resistors (R1 and R2) and a red LED. The battery is ideal and maintains a constant voltage of +9.0 V (i.e., V1 - V2 = +9.0 V). An ideal voltmeter is connected across the LED and R2. At point a, is the direction of the current to the right, to the left, or equal to zero? To the right. To the left. Equal to zero. There is not enough information to answer the question. Rank, from largest to smallest, the absolute values of the currents at points a, b, c, and d. If any of the currents are zero, state so explicitly. a = b = c = d = 0. a = b = c = d > 0. a = b > c = d = 0. a > b > c = d > 0. a = c = d > b = 0. b = c = d > a = 0. c = d > b > a > 0. There is not enough information to answer the question. Explain your reasoning. Rank, from largest to smallest, the absolute values of the voltages across R1, R2, and the LED. If any of the voltages are zero, state so explicitly. R1 = R2 = LED = 0. R1 = R2 = LED > 0. R1 = R2 > LED > 0. R1 > R2 = LED = 0. R1 > LED > R2 > 0. LED > R1 = R2 = 0. There is not enough information to answer the question. Explain your reasoning. What, if anything, can be said about the absolute value of the reading on the voltmeter? 0.0 V. 9.0 V . The absolute value of the voltmeter reading will be greater than 0.0 V and less than 9.0 V. There is not enough information to answer the question.

Explanation / Answer

direction of current at a is towards right..... a > b > c = d > 0 ............ there will be a current in diode (reverse saturation current) hence c=d , voltmeter will also draw a current ...from circuit.................... .current at a = diode current + voltmeter current.......... ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: R1 > LED > R2 > 0 ................... because current at a is more than diode , hence voltage across R1 is more than R2.... , voltage across LED is more than R2 because current through LED is in micro ampere ,,, :::::::::::::::::::::::::::::::::::::::::::: voltmeter reading = 9 -i*R2 ..... , where i is current through R2 ....

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