A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of $$10\Omega$$ and a resistance $$R$$, to a 100 V mains as shown in figure. For the heater to operate at 62.5 W, the value of $$R$$ should be _____ $$\Omega$$

Step 1: Calculate the resistance of the heater($$R_H$$)

Using the rated power and voltage:

$$ R_H = \frac{V_{rated}^2}{P_{rated}} $$

$$ R_H = \frac{100^2}{1000} $$

$$ R_H = 10 \Omega $$

Step 2: Calculate the voltage across the heater when operating at 62.5 W

Let the new voltage across the heater be $$V_H$$.

$$ P_{new} = \frac{V_H^2}{R_H} $$

$$ 62.5 = \frac{V_H^2}{10} $$

$$ V_H^2 = 625 $$

$$ V_H = 25 \text{ V} $$

Step 3: Determine the total current in the circuit ($$I_{total}$$)

The heater and the resistor $$R$$ are in parallel. This combination is in series with the $$10 \Omega$$ resistor.

The voltage across the parallel combination (between points B and C) is $$V_{BC} = V_H = 25 \text{ V}$$.

The total voltage of the source is $$100 \text{ V}$$.

The voltage across the $$10 \Omega$$ series resistor is:

$$ V_{series} = 100 - V_{BC} $$

$$ V_{series} = 100 - 25 = 75 \text{ V} $$

Now, find the total current flowing through the $$10 \Omega$$ series resistor:

$$ I_{total} = \frac{V_{series}}{10} $$

$$ I_{total} = \frac{75}{10} = 7.5 \text{ A} $$

Step 4: Determine the current through the unknown resistor $$R$$

The total current splits at node B into the heater and resistor $$R$$.

First, find the current through the heater ($$I_H$$):

$$ I_H = \frac{V_H}{R_H} = \frac{25}{10} = 2.5 \text{ A} $$

Using Kirchhoff's Current Law ($$I_{total} = I_H + I_R$$):

$$ I_R = I_{total} - I_H $$

$$ I_R = 7.5 - 2.5 = 5 \text{ A} $$

Step 5: Calculate the value of $$R$$

The voltage across $$R$$ is $$V_{BC} = 25 \text{ V}$$, and the current through it is $$I_R = 5 \text{ A}$$.

$$ R = \frac{V_{BC}}{I_R} $$

$$ R = \frac{25}{5} $$

$$ R = 5 \Omega $$