Question

Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in parallel to the first two. The total power dissipated in this circuit is: a. 2P/3 b. 3Pc. 2Pd. 3P/2e. P

Answer 1

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit  such that

$P=VI=V^2/R$

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

$Current2=\frac{V}{2R}$

Power2=Voltage * Current2

$Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2$

$TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}$