A lightbulb with a resistance of 323 Ohm is connected to a power supply that pro

Question

A lightbulb with a resistance of 323 Ohm is connected to a power supply that provides 220 V by two connecting wires, each with a resistance of 12 Ohm. What is the equivalent resistance of this circuit? (347 Ohm) What is the current through the circuit? [.634 A] What is the magnitude of the potential difference between point A and point B? (This is the voltage across wire 1.) [7.6 V] What is the magnitude of the potential difference between point B and point C? (This is the voltage across the bulb.) [204.8 V] Continuing #3. How much power is "lost" in wire 1? [4.8 W] How much power is used by the bulb? [129.8 W] Continuing #4, you want your bulb to be brighter (use more power, less power lost in the wires) so you switch the connecting wires for ones with less resistance. Assuming you're using the same power supply, what resistance should you make each wire (keep R_wise 1 = R_wise 2) so that the bulb is using 140 W? (5.6 Ohm]

Explanation / Answer

resistance of bulb Rb = 323 ohm

and of wires Rw1 = Rw2 = 12

let Rw = Rw1 + Rw2 = 12 + 12 = 24 (both in series)

a) equivalent resistance = 323 + 24 = 347 ohms

b) current i = 220/347 = 0.634 A

c) Va - 0.634*12 = Vb

Va - Vb = 7.6V

d) similarly Vb - Vc = 323*0.634 = 204.8

4) a) power in w1 = i^2 * Rw1 = 0.634*0.634*12 = 4.8 W

b)power lost in bulb = 0.634*0.634 * 323 = 129.8 W

5) power in bulb = 140W

so, i^2 * 323 = 140

i = 0.658 A

voltage lost in the bulb is V^2 = 140 * 323 = 45220

Vb = 212.6 V

remaining voltage drop is due to wires = 220 - 212.6 = 7.4 V

total resistance of 2 wires = 7.4/0.658 = 11.2 ohms

resistance of 1 wire = 5.6 ohms

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