Question

Two bulbs of 500 W and 200 W are rated to operate on 220 V. The ratio of heat produced in the combination when they are connected in series and in parallel

• A. $5/2,2/5$
• B. $5/2,5/2$
• C. $2/5,5/2$
• D. $2/5,2/5$

Answer 1

The correct option is C $2/5,5/2$

Let, $P_1$ and $P_2$ be the power of the two bulbs and $R_1$ and

$R_2$ be the resistances of two bulbs.

Power of the bulb 1 is

$P_1=\frac{V^2}{R_1}$

$R_1=\frac{V^2}{P_1}$

$R_1=\frac{(220{)}^2}{500}$

$R_1=96.8\Omega$

Similarly, power of the bulb 2 is

$P_2=\frac{V^2}{R_2}$

$R_2=\frac{V^2}{P_2}$

$R_2=\frac{(220{)}^2}{200}$

$R_2=242\Omega$

A] When bulbs are connected in series, current through both bulbs is same.

The equivalent resistance in series is

$R_s=R_1+R_2$

$R_s=96.8+242$

$R_s=338.8\Omega$

Hence, the current through the circuit is

$I_s=\frac{V}{R_s}$

$I_s=\frac{220}{338.8}$

$I_s=0.65A$

Heat produced in bulbs are

$H_1=I_s^{2}R$

$H_1=\left(0.65{)}^2\right(96.8)$

$H_1=40.89J$

and

$H_2=I_s^{2}R$

$H_2=\left(0.65{)}^2\right(242)$

$H_2=102.24J$

Therefore,

$\frac{H_1}{H_2}=\frac{40.89}{102.24}\approx \frac{2}{5}$

B] When bulbs are connected in parallel, voltage across the bulbs are same

Hence, the current through the bulb 1 is

$I_1=\frac{V}{R_1}$

$I_1=\frac{220}{96.8}$

$I_1=2.272A$

Heat produced in bulb 1 is

$H_1=I_1^{2}R$

$H_1=\left(2.272{)}^2\right(96.8)$

$H_1=508.97J$

and the current through the bulb 1 is

$I_1=\frac{V}{R_2}$

$I_1=\frac{220}{242}$

$I_1=0.09A$

Heat produced in bulb 2 is

$H_2=I_2^{2}R$

$H_2=\left(0.09{)}^2\right(242)$

$H_2=1.96J$

Therefore,

$\frac{H_1}{H_2}=\frac{508.97}{1.96}\approx \frac{50.9}{19.6}\approx \frac{5}{2}$