What will be the resistance (in ohms) of bulb A for the circuit given below?

• A. 1.33
• B. 4.65
• C. 3.3
• D. 2.36

Answer: (C)

The power can be defined as
P = I2R
Let the Power dissipated by Bulb A be
P = I2RA = 100 = I2RA
And power dissipated by Bulb B be
P = I2RB = 10 = I2RB
As we Know that  the current flow in the series circuit is same
RA ⁄ 100 = RB ⁄ 10
RA ⁄ RB = 10
or
RA = 10RB
Power can also be defined as the
P = V2/R
Total Power consumption = 100 Watt + 10 Watt = 110 Watt
Applied voltage = 20 Volt
∴ 110 = 202 ⁄ R
or
R = 202 ⁄ 110 = 40 ⁄ 11
Now in series connection, the equivalent resistance is the sum of the individual resistance
∴ R = RA + RB
40 ⁄ 11 = 10RB + RB
RB = 0.33 Ω
Hence RA = 10RB
= 10 × 0.33 =3.3
RA = 3.3Ω