#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# I have posted my question as an attachement.  Kindly refer to that .

M H Rashid
35 Points
4 years ago
First we need to find out the resistance of the bulbs, then the equivalent resistance on series combination and finally the power consumption as required in the problem.
$\fn_jvn P=V I=\frac{V^{2}}{R}$
$\fn_jvn \therefore R=\frac{V^{2}}{P}$
Given: V1=V=200 V, P1= 60 W
V2=V=200 V, P2=100 W
$\fn_jvn \therefore R_{1}=\frac{V^{2}}{P_{1}}$
and
$\fn_jvn R_{2}=\frac{V^{2}}{P_{2}}$
Therefore the equivalent resistance is
$\fn_jvn R=R_{1}+R_{2}=V^{2}\left \{ \frac{1}{P_{1}}+\frac{1}{P_{2}} \right \}=V^{2}\left \{ \frac{P_{1}+P_{2}}{P_{1}P_{2}} \right \}$
Hence the Power consumption would be
$\fn_jvn P=\frac{V^{2}}{R}=\frac{P_{1}P_{2}}{P_{1}+P_{2}} =\frac{60\times 100}{60+100}=\frac{75}{2}=37.5 W$