Five identical resistors, each of resistance R, are connected in series to a bat

Question

Five identical resistors, each of resistance R, are connected in series to a battery, and a total current of 7.88 A flows in the circuit. Now, another resistor of resistance R is added to the circuit, in series with the rest of the resistors. Find the new total current in the circuit. I = A In the original circuit: suppose another resistor of resistance R is added to the circuit, in parallel with the battery. Find the new total current in the circuit. I = A Five identical resistors, each of resistance R, are connected in parallel to a battery, and a total current of 7.88 A flows in the circuit. Now, another resistor of resistance R is added to the circuit, in parallel with the rest of the resistors. Find the new total current in the circuit. 1=

Explanation / Answer

a) Total resistance of 5 series resistors = 5R

Current = V/5R = 7.88 =======> V/R = 39.4

i) After adding one resistor in series, net resistance = 6R ============> Current = I = V/6R = 39.4/6 = 6.57 A

ii) After adding one resistor in parallel, net resistance = 0.833R ============> Current = I = V/0.833R = 47.3 A

b) Total resistance of 5 parallel resistors = R/5

Current = V/(R/5) = 7.88 =======> V/R = 1.576

i) After adding one resistor in parallel, net resistance = 0.167R ======> Current = I = V/0.167R = 9.44 A

ii) After adding one resistor in series, net resistance = 1.2R ============> Current = I = V/1.2R = 1.31 A

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