Three identical light bulbs A, B and C are connected so that A and B in series are in parallel with C?

Question

All bulbs are marked 240 V, 100 W. A 240 V ac supply is then connected across C.

a) Find the bulb resistance when operating Normally.
b) Find the currents in A, B and C.
c) Which bulb will be least bright?
d) What would be the effect of removing bulb C from its holder?

Answer 1

(a) Using ohms law and the definition of a watt

you get R=(E squared)/W so R=(240x240)/100 = 576 ohms

(b) The current in a bulb is I = E/R

for C you get I=240/576 = 0.42 amps
for A or for B which see half of the 240 volts
you get I=120/576 = 0.21 amps
(c) The bulbs A and B are each dimmer than bulb C
(d) C would go out and A and B would be the same brightness.

Answer 2

Wire bulbs a and b in series. One terminal from a is wired to one terminal of C, and the open terminal from B is wired to the other terminal of C. Power is applied across bulb C.

Bulb C will be the brightest, and A and B will be of equal brightness, less than C.

Opening bulb C will have no effect on A or B.

The formula for Power is I^2 * R and Voltage * current. a 100 Watt Bulb at 100 Volts draws 1 Ampere. You should be able to do the algebra.

Answer 3

problematic task. seek from bing and yahoo. that might help!