Four identical bulbs, each with a filament resistance of 53 ?, are connected to

Question

Four identical bulbs, each with a filament resistance of 53 ?, are connected to a 45-V battery as shown in the diagram below. S1, S2, and S3 represent switches.

(d) If S1 and S3 are closed while S2 is open, what is the current through each bulb?

This was a multi part problem, and i understand the first 3 parts, however due to the configuration of this last part, im not sure how to set up the problem. i calculated 3+4 to be a series resistance of 106ohm, then bulb2 and bulb34 to be in parallel with eachother, which is then in series again with bulb 1.

i calculated the total resistance to be 91.3 ohm, which means a total current of 49.28amp.

this is where i get confused... i was taking my known amperage and the resistance of each resistor and doing I*R to get voltage... since once i have each resistors voltage, i can then re-solve for the lone resostors current... however i get the following diagram:

i feel like im doing something wrong here.... that would only give me .27 volts to play with for the two resistors in series (bulbs 3+4). assuming thats wright, i still am not sure what to do.

Explanation / Answer

When s1 and s3 are kept cosed and s2 is kept opned, circuit will be looked like as below,

Applying KVL in loop1,

V - I2*R - I1*R =0

45 – 53I1 – 53I2 = 0 ---------(1)

Applying KVL in loop2,

I2*R = I3*(2R)

I2/I3 = 2R/R = 2 -------------(2)

At nde point a apply KCL,

I1= I2 + I3

By (2)

I1= I2 + 2I2 = 3I2   ---------(3)

Pu (3) in (1)

45 – 53(3I2) – 53I2 = 0

45 – 4(53I2) = 0

I2 =0.21 A

I3= 2I/22 = 0.21/2 = 0.105A

I1=3I2 = 3*0.21 = 0.63A

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