Question
A) The current in a 47- resistoris 0.12 A. This resistor is in series with a 28-resistor, and the series combination is connected across abattery. What is the battery voltage? B) What resistance must beplaced in parallel with a 155- resistor to make theequivalent resistance 115 ? Please help! I don't get this at all. I will rate youlifesaver immediately! A) The current in a 47- resistoris 0.12 A. This resistor is in series with a 28-resistor, and the series combination is connected across abattery. What is the battery voltage? B) What resistance must beplaced in parallel with a 155- resistor to make theequivalent resistance 115 ? Please help! I don't get this at all. I will rate youlifesaver immediately!
Explanation / Answer
A)The current in a 47- resistor is 0.12 A. The 47- resistor is in series with a 28-resistor.Therefore,the total resistance of the two series resistorsis R = R1 + R2 or R = 47 + 28 = 75- The battery voltage is V = I * R Here,I = 0.12 A or V = 0.12 * 75 = 9 V B)Let the resistor that should be placed in parallel with a155- resistor to make the equivalent resistance 115 beR.Therefore,we get (1/R) + (1/155) = (1/115) or (1/R) = (1/115) - (1/155) = (8/3565) or R = (3565/8) = 445.625- or R = 47 + 28 = 75- The battery voltage is V = I * R Here,I = 0.12 A or V = 0.12 * 75 = 9 V B)Let the resistor that should be placed in parallel with a155- resistor to make the equivalent resistance 115 beR.Therefore,we get (1/R) + (1/155) = (1/115) or (1/R) = (1/115) - (1/155) = (8/3565) or R = (3565/8) = 445.625-