When three equal resistors are connected in series across a source of emf, the total resistance of the circuit, R_total, is equal to the sum of the individual resistances, R. Let's assume the resistance of each resistor is R.

1. Calculation of Total Resistance in Series:
In a series circuit, the total resistance is the sum of the individual resistances. Therefore,
R_total = R + R + R = 3R

2. Calculation of Power Dissipated in Series:
The power dissipated in a resistor can be calculated using the formula P = I^2 * R, where P is power, I is current, and R is resistance. In a series circuit, the current passing through each resistor is the same.

Given that the total power dissipated in the circuit is 10 watts, we can write the equation as:
10 = I^2 * 3R

3. Calculation of Power Dissipated in Parallel:
When the resistors are connected in parallel, the voltage across each resistor is the same, but the current divides among the resistors. Let's assume the current passing through each resistor is I_parallel, and the total current passing through the parallel combination is I_total.

Using Ohm's Law, we can write the equation for the total current:
I_total = V / R_total

Since the resistors are equal, we can write the equation for the current passing through each resistor in parallel:
I_parallel = I_total / 3

The power dissipated in each resistor can be calculated using the formula P = I^2 * R:
P_parallel = (I_parallel)^2 * R

Substituting the values of I_parallel and R, we get:
P_parallel = (I_total / 3)^2 * R

4. Calculation of Power Dissipated in Parallel:
To find the power dissipated in the parallel combination, we need to substitute the value of I_total from the equation of total current:
P_parallel = ((V / R_total) / 3)^2 * R

Simplifying the equation gives:
P_parallel = (V^2 / (9 * R)) * R

Using the formula for power in a series circuit, we know that P = I^2 * R. Since the total resistance in the parallel combination is R, we can write the equation as:
P_parallel = (V^2 / (9 * R)) * R = (V^2 / 9)

From the equation, it is clear that the power dissipated in the parallel combination is independent of resistance. Therefore, the power dissipated in the resistors when connected in parallel would be equal to the power dissipated in the series combination, which is 10 watts. Hence, the correct answer is option C - 90 watts.