A dc source of emf $$E_1 = 100$$ V and internal resistance $$r = 0.5\Omega$$, a storage battery of emf $$E_2 = 90$$ V and an external resistance R are connected as shown in figure. For what value of R no current will pass through the battery?

We are given a DC source with emf $$ E_1 = 100 $$ V and internal resistance $$ r = 0.5 \, \Omega $$, a storage battery with emf $$ E_2 = 90 $$ V, and an external resistance $$ R $$. The goal is to find the value of $$ R $$ such that no current passes through the battery. This means the battery $$ E_2 $$ is effectively inactive, and the voltage across its terminals must equal its emf, $$ 90 $$ V.

In the circuit, the DC source $$ E_1 $$ with internal resistance $$ r $$ is connected in series with the external resistance $$ R $$. The battery $$ E_2 $$ is connected in parallel with $$ R $$. Therefore, the voltage across $$ R $$ must equal $$ E_2 $$ for no current to flow through the battery. So, we set the voltage across $$ R $$ equal to $$ 90 $$ V.

The total resistance in the series circuit is $$ r + R = 0.5 + R $$ ohms. The current $$ I $$ flowing through the circuit is given by Ohm's law:

$$ I = \frac{E_1}{r + R} = \frac{100}{0.5 + R} $$

The voltage across $$ R $$ is $$ V_R = I \times R $$. Setting this equal to $$ E_2 $$:

$$ V_R = \frac{100}{0.5 + R} \times R = 90 $$

Now, solve for $$ R $$:

$$ \frac{100R}{0.5 + R} = 90 $$

Multiply both sides by $$ 0.5 + R $$:

$$ 100R = 90 \times (0.5 + R) $$

Calculate $$ 90 \times 0.5 = 45 $$:

$$ 100R = 45 + 90R $$

Subtract $$ 90R $$ from both sides:

$$ 100R - 90R = 45 $$

$$ 10R = 45 $$

Divide both sides by 10:

$$ R = \frac{45}{10} = 4.5 \, \Omega $$

Thus, when $$ R = 4.5 \, \Omega $$, no current flows through the battery. Comparing with the options:

A. $$ 5.5 \, \Omega $$

B. $$ 3.5 \, \Omega $$

C. $$ 4.5 \, \Omega $$

D. $$ 2.5 \, \Omega $$

Hence, the correct answer is Option C.