Question

Two appliances of rating 200 watt-250 volts and 100 watt-250 volts
are joined in series to a 250 volts supply. Total power consumed in the
circuit is​

Answer 1

Answer:

For bulb 1

P = 200 watt

V = 250 V

I = P/V

I = 200/250

I = 0.8 A

R = V/I

R = 250/0.8

R = 312.5 ohm

For bulb 2

P = 100 watt

V = 250 V

I = P/V

I = 100/250

I = 0.4 A

R = V/I

R= 250/0.4

R = 625 ohm

Total resistance in circuit = 625 + 312.5

= 937.5 ohm

Total power consumed, P = V²/R

= 250 × 250/937.5

= 66.66 watt or 67 watt

PLEASE MARK IT BRAINILIEST

Answer 2

Given,

Rated power of the first bulb ($P_{1}$) = 200 watt

Rated power of the second bulb ($P_{2}$) = 100 watt

Voltage (V) = 250 Volts

To Find,

Total power consumed in the  circuit

Solution,

We know that the power dissipated in a resistor is given by

     $P = \frac{V^{2}}{R}$

Therefore, Resistance for first bulb is,

     $R_1 = \frac{V^2}{P_1}$

Substituting the values we get,

     $R_1 = \frac{250^2}{200}$

     $R_1 = \frac{625}{2} ohm$

Resistance for second bulb is,

     $R_2 = \frac{V^2}{P_2}$

Substituting the values we get,

     $R_2 = \frac{250^2}{100}$

     $R_2 = 625 ohm$

Total resistance of the circuit when it is connected in series is given by,

     $R = R_1+R_2$

     $R = \frac{625}{2} + 625$

     $R = \frac{1875}{2} ohm$

Therefore, total power consumed in a circuit is,

    $P = \frac{V^{2}}{R}$

Substituting the values we get,

     $P = \frac{250^2}{1875/2}$

     $P = 66.67 W$

The total power consumed in the  circuit is​ 66.67 Watt.