If a person has five resistors each of value $\frac{1}{5} \Omega$, then the maximum resistance he can obtain by connecting them is

(A) $1 \Omega$

(B) $5 \Omega$

(C) $10 \Omega$

(D) $25 \Omega$

OR

The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become

First Part Analysis:

To obtain the maximum resistance using a given set of resistors, they must be connected in series.

The formula for the equivalent resistance $R_s$ of resistors connected in series is:

$$R_s = R_1 + R_2 + R_3 + R_4 + R_5$$

Given that there are five resistors, each of value $R = \frac{1}{5} \Omega$:

$$R_s = 5 \times \frac{1}{5} \Omega = 1 \Omega$$

Thus, the maximum resistance is $1 \Omega$, which corresponds to option (A) of the first part.

OR Part Analysis:

According to Joule's law of heating, the heat produced or power consumed in a resistor connected across a constant voltage source $V$ is given by:

$$P = \frac{V^2}{R}$$

If the resistance is reduced to half of its initial value ($R' = \frac{R}{2}$) while the other parameters (like the potential difference $V$ of the source) remain unchanged:

$$P' = \frac{V^2}{R'} = \frac{V^2}{R/2} = 2 \left(\frac{V^2}{R}\right) = 2P$$

Therefore, the heating effect in the resistor will become two times its initial value. This corresponds to option (A).

Explanation of Distractors for the OR Part:

• Option B (half): This would be correct if the current $I$ through the resistor was kept constant ($P = I^2 R$), but in a standard circuit, the voltage source remains constant, causing the current to double when resistance is halved.
• Option C (one-fourth): This incorrect factor arises from misapplying formulas or incorrect algebraic manipulation.
• Option D (four times): This would occur if the current was doubled while keeping the resistance constant, which is not the case here.