Question

Three electric bulbs rated $40,60$ and $100$ watts respectively are connected in parallel with $230$ volts d.c mains. The current (in amp) through the $60W$ bulb and the effective resistance (in ohm) of the circuit will be respectively, approximately

• A. $0.23A$
• B. $329\Omega$
• C. $0.26A$
• D. $265\Omega$

Answer 1

Answer: (C), (D)

The correct options are
C $0.26A$
D $265\Omega$

In parallel connections voltages in all branches are same and as :

Power = $\frac{V^2}{R}$

We are increasing power that means increasing resistance, as the current distribution follows inverse law, the current in other branch will increase as a result the heater will produce more heat.

$H=I^{2}R$