Beam me up, Date 4.I have 4 circuits. Each circuit contains a 12 V battery (or 1

Question

Beam me up, Date 4.I have 4 circuits. Each circuit contains a 12 V battery (or 12 V worth of batteries) and either 2 or 3 light bulbs, each having a resistance of 2 ohms. Circuit I: Battery is connected to two bulbs in series. Circuit 2: Battery is connected to three bulbs in series. Circuit 3: Battery is connected to two bulbs in parallel. Circuit 4: Battery is connected to three bulbs, two are in parallel and this parallel structure is in series with the 3 RS 2 (a) Rank the above circuits in order from greatest to least amount of current flowing out of batt Cirund 3 (b) Rank the above circuits in order from greatest to least amount of the smallest current flowing through any part of the circuit. [For example, if circuit 1 has the following currents flowing through the wires: 2 A, -3 A, 4 A; and circuit 2 has the following: 3 A, -5 A, 1 A; then I would look at the 2 A from circuit 1 and compare it to the 1 of circuit 2. I would rank them as Circuit 1, Circuit 2.] A Cirtnt runt 2 uww

Explanation / Answer

a)

as per ohm's law

i = V/R so greater the resistance of the circuit , smaller will be the current

circuit 1 :

r = resistance of each bulb = 2 ohms

R1 = combination of resistance of 2 bulbs in series = r + r = 2r = 2 x 2 = 4 ohm

circuit 2 :

r = resistance of each bulb = 2 ohms

R2 = combination of resistance of 3 bulbs in series = r + r + r = 3r = 3 x 2 = 6 ohm

circuit 3 :

r = resistance of each bolb = 2 ohms

R3 = combination of resistance of 2 bulbs in parallel= r.r/(r + r) = r/2 = 2/2 = 1 ohm

circuit 4 :

r = resistance of each bulb = 2 ohms

R4 = combination of resistance of 2 bulbs in parallel and one in series = r.r/(r + r) + r = r/2 + r = 1.5 ohm

so rank of circuits with greatest to least current will be

ciruit 3 >circuit 4 >circuit 1 > circuit 2

b)

circuit 1 :

r = resistance of each bulb = 2 ohms

R1 = combination of resistance of 2 bulbs in series = r + r = 2r = 2 x 2 = 4 ohm

i1 = current = V/R1 = 12/4 = 3 A

circuit 2 :

r = resistance of each bulb = 2 ohms

R2 = combination of resistance of 3 bulbs in series = r + r + r = 3r = 3 x 2 = 6 ohm

i2 = current = V/R2 = 12/6 = 2 A

circuit 3 :

r = resistance of each bolb = 2 ohms

R3 = combination of resistance of 2 bulbs in parallel= r.r/(r + r) = r/2 = 2/2 = 1 ohm

i = current in each parallel resistor = V/r = 12/2 = 6 A

circuit 4 :

r = resistance of each bulb = 2 ohms

R4 = combination of resistance of 2 bulbs in parallel and one in series = r.r/(r + r) + r = r/2 + r = 1.5 ohm

i4 = current in series resistor = V/R4 = 12/1.5 = 8 A

i' = current in each parallel resistor = (12 - 8)/2 = 2 A

so rank of circuits with greatest to least current will be

ciruit 3 >circuit 4 >circuit 1 > circuit 2

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