(a) Why does an electric bulb become dim when an electric heater in a parallel circuit is switched ON?

(b) How would you connect three resistors, each of resistance $8\,\Omega$, so that the equivalent resistance of the combination is $12\,\Omega$? Draw the diagram of the combination and justify your answer.

(a) Why the bulb dims:

In a household circuit, the bulb and the heater are connected in parallel across the 220 V supply, so ideally the potential difference (V) across the bulb should stay the same (220 V) whether or not the heater is switched on — the NCERT text states that in a parallel circuit, the potential difference across each component is the same, and each gadget draws current according to its own resistance (Section 11.6.2).

However, no real source of supply (a cell, battery, or the mains transformer/generator feeding the house) is an ideal source of zero internal resistance; the connecting wires also have some resistance. When the heater (a low-resistance, high-current appliance) is switched ON in parallel with the bulb, the total current drawn from the source increases (Eq. 11.15, $I = I_1+I_2+I_3$). This larger total current flowing through the small internal/line resistance produces a larger voltage drop across that internal resistance ($V=IR$). Consequently, the voltage actually delivered to the parallel combination (and hence across the bulb) drops slightly below 220 V. Since the power delivered to the bulb filament falls as $P=V^2/R$, the bulb glows dimmer.

(b) Combination for $12\,\Omega$:

Given three resistors, each $R=8\,\Omega$.

Step 1 — Put two of them in parallel:

$$\frac{1}{R_p}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4} \implies R_p = 4\,\Omega$$

Step 2 — Connect this parallel pair in series with the third $8\,\Omega$ resistor:

$$R_{\text{eq}} = R_p + R_3 = 4\,\Omega + 8\,\Omega = 12\,\Omega$$

Circuit description: Between terminals X and Y, two $8\,\Omega$ resistors are joined side by side (both ends common) to form the parallel block of $4\,\Omega$; one end of this block is then joined in series (end to end, single path) to the third $8\,\Omega$ resistor, and the free ends of this series combination are connected to the battery through a plug key. This matches Eq. (11.14) for series ($R_s = R_1+R_2$) and Eq. (11.18) for parallel resistors, exactly as developed in the NCERT chapter, and gives the required equivalent resistance of $12\,\Omega$.