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

Question

Five identical resistors, each of resistance R, are connected in series to a battery, and a total current of 4.4 A flows in the circuit.
i. 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 = 3.67   A
ii. 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

b) Five identical resistors, each of resistance R, are connected in parallel to a battery, and a total current of 4.4 A flows in the circuit.
i. 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.
I =   A
ii. In the original circuit: suppose another resistor of resistance R is added to the circuit, in series with the battery. Find the new total current in the circuit. I =   A

Explanation / Answer

(a)

The resistance will increase by a factor of 6/5, so the current will decrease to 5/6 of its initial value.
4.4 A x (5/6) = 3.67 A
(b)

The resistance will decrease to 5/6 of its initial value, so the current will increase by a factor of 6/5.
4.4 A x (6/5) = 5.28 A

(c)

The easiest way to do this is to assign a voltage for the source, Vs. Assigning 10V, the total resistance of the five resistors is 10 / 4.4 = 2.272 ohms. The resistance of each resistor is 2.272 / 5 = 0.4545 ohm. The equivalent resistance, R(eq), of 2.272 ohm in parallel with 0.4545 ohm, is (2.272*0.4545) / (2.272 + 0.4545) = 0.3787 ohm. Current from the source is

I = 10 / 0.3787 = 26.41 amps

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