A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1 $$\Omega$$. The battery terminals are connected to external resistance $$R$$. The minimum value of $$R$$, so that a current passes through the battery to charge it is:

The circuit consists of a DC main supply of 220 V connected across a storage battery of 200 V through a 1 Ω resistance. The battery terminals are also connected to an external resistance $$R$$. We need to find the minimum value of $$R$$ such that current flows through the battery to charge it, meaning current enters the positive terminal of the battery.

First, we assume the battery has no internal resistance, as it is not specified in the problem. Therefore, the voltage across the battery terminals is always equal to its EMF, which is 200 V. Let the common negative terminal be point C, and the positive terminal of the supply be point A. The supply is connected to a 1 Ω resistor, and the other end of this resistor is connected to the positive terminal of the battery (point B) and one end of $$R$$. The negative terminal of the battery and the other end of $$R$$ are connected to point C.

Since the voltage at point B with respect to point C is the same as the battery voltage, $$V_{BC} = 200$$ V. The supply voltage is 220 V between A and C, so $$V_{AC} = 220$$ V. The voltage drop across the 1 Ω resistor (between A and B) is $$V_{AB} = V_{AC} - V_{BC} = 220 - 200 = 20$$ V.

The current through the 1 Ω resistor, flowing from A to B, is given by Ohm's law: $$I = \frac{V_{AB}}{1 \Omega} = \frac{20}{1} = 20$$ A.

At junction B, this current $$I$$ splits into two parts: $$I_b$$ through the battery and $$I_R$$ through the external resistance $$R$$. Applying Kirchhoff's current law: $$I = I_b + I_R$$, so $$20 = I_b + I_R$$.

The voltage across $$R$$ is the same as $$V_{BC} = 200$$ V, so by Ohm's law, $$I_R = \frac{V_{BC}}{R} = \frac{200}{R}$$. Substituting this in: $$20 = I_b + \frac{200}{R}$$.

For the battery to charge, current must enter its positive terminal, so $$I_b > 0$$. Therefore, $$20 - \frac{200}{R} > 0$$. Solving this inequality:

$$20 > \frac{200}{R}$$

Since $$R$$ is positive, multiply both sides by $$R$$: $$20R > 200$$.

Divide both sides by 20: $$R > 10 \Omega$$.

Thus, $$R$$ must be greater than 10 Ω for charging to occur. The minimum value of $$R$$ in the given options that satisfies this condition is 11 Ω.

Hence, the correct answer is Option C.